SELECTION OF TERMS IN G.P.
Sometimes it is required to select a finite number of terms in G.P. It is always convenient if we select the terms in the following manner :
No. of Terms | Terms | Common Ratio |
3 4 5 | r r2 r |
If the product of the numbers is not given, then the numbers are taken as a, ar, ar2, ar3, ...
TRY OUT THE FOLLOWING
1. If the sum of three numbers in G.P. is 38 and their product is 1728, find them.
2. Find the three numbers in G.P. is whose sum is 13 and the sum of whose squares is 91.
3. Find four numbers in G.P. whose sum is 85 and product is 4096.
4. Three numbers are in G.P. whose sum is 70. If the extremes be each multiplied by 4 and the means by 5, they will be in A.P. Find the numbers.
5. Find four numbers in G.P. in which the third term is greater than the first by 9 and the second term is greater than the fourth by 18.
6. The product of first three terms of a G.P. is 1000. If 6 is added to it's second term and 7 added to its third term, the terms become in A.P. Find the G.P.
ANSWERS
1.8, 12, 18, or 18, 12, 8 2.1, 3, 9 or 9, 3, 1 3.1, 4, 16, 64 or 64, 16, 4, 14.10, 20, 40, or 40, 20, 10 5. 3, – 6, 12, – 24 6. 5, 10, 20,... or 20, 10, 5... ˜
SUMTON TERMS OF AN ARITHMETIC PROGRESSION
The sum Sn of n terms of an arithmetic progression with first term 'a' and common difference 'd' is
OR
Where ℓ = last term.
Remark-1 : In the formula there are four quantities viz. Sn, a, n and d. If anythree of these are known, the fourth can be determined. Sometimes, two of these quantities are given. In such a case, remaining two quantities are provided by some other relation.
Remark-2 : If the sum Sn of n terms of a sequence is given, then nth term an of the sequence can be determined by using the following formula : an = Sn – S n –1 i.e., the nth term of an AP is the difference of the sum to first n terms and the sum to first (n –1) terms of it.
Ex.14 Find the sum of the AP:
⇒
⇒
⇒
So, the sum of the first 11 terms of the given AP is 33/20
Ex.15 Find the sum : 34 + 32 + 30 + .... + 10
Sol. 34 + 32 + 30 + .... + 10
This is an AP
Here, a = 34 d = 32 – 34 = – 2
ℓ = 10
Let the number of terms of the AP be n.
We know that
an = a + (n – 1)d
⇒ 10 = 34 + (n – 1) (–2)
⇒ (n – 1) (– 2) = – 24
Again, we know that
Hence, the required sum is 286.
Ex.16 Find the sum of all natural numbers between 100 and 200 which are divisible by 4.
Sol. All natural numbers between 100 and 200 which are divisible by 4 are
104, 108, 112, 116,...,196
Here, a1 = 104
a2 = 108
a3 = 112
∴ a2 – a1 = 108 – 104 = 4 a3 – a2 = 112 – 108 = 4 a4 – a3 = 116 – 112 = 4 ·
∵ a2 – a1 = a3 – a2 = a4 – a3 = ..... (= 4 each)
∴ This sequence is an arithmetic progression whose common difference is 4.
Here,a = 104
d = 4
ℓ = 196
Let the number of terms be n. Then ℓ= a + (n – 1)d
⇒ 196 = 104 + (n – 1)4
⇒ 196 – 104 = (n – 1)4 ⇒ 92 = (n – 1) 4
⇒ (n – 1) 4 = 92 ⇒ n – 1 = 92/4
⇒ n – 1 = 23 ⇒ n = 23 + 1 ⇒ n = 24
Again, we know that
⇒ S24 = (104 + 196)
= (12) (300) = 3600
Hence, the required sum is 3600.
Ex.17 Find the number of terms of the AP 54, 51, 48,...so that their sum is 513.
Sol. The given AP is 54, 51, 48,...
Here, a = 54, d = 51– 54 = – 3 Let the sum of n terms of this AP be 513.
We know that
⇒ 1026 = 111n – 3n2 ⇒ 3n2 – 111n + 1026 = 0
⇒ n2 – 37n + 342 = 0 [Dividing throughout by 3]
⇒ n2 – 18n – 19n + 342 = 0 ⇒ n(n – 18) – 19(n – 18) = 0
⇒ (n – 18) (n – 19) = 0 ⇒ n – 18 = 0 or n – 19 = 0
⇒ n = 18, 19 Hence, the sum of 18 terms or 19 terms of the given AP is 513.
Note : Actually 19th term
= a19
= a + (19 – 1) d [∵an = a + (n – 1)d]
= a + 18d = 54 + 18 (– 3)
= 54 – 54 = 0
Ex.18 Find the AP whose sum to n terms is 2n2 + n.
Sol. Here, Sn = 2n2 + n (Given)
Put n = 1, 2, 3, 4,..., in succession, we get
S1 = 2(1)2 + 1 = 2 + 1 = 3
S2 = 2(2)2 + 2 = 8 + 2 = 10
S3 = 2(3)2 + 3 = 18 + 3 = 21
S4 = 2(4)2 + 4 = 32 + 4 = 36
and so on.
∴a1 = S1 = 3
a2 = S2 – S1 = 10 – 3 = 7
a3 = S3 – S2 = 21 – 10 = 11
a4 = S4 – S3 = 36 – 21 = 15 and so on.
Hence, the required AP is 3, 7, 11, 15,...
Ex.19 200 logs are stacked in the following manner : 20 logs in the bottom row, 19 in the next row, 18 in the row next to it and so on (see figure). In how many rows are the 200 logs placed and how many logs are in the top row?
Sol. The number of logs in the bottom row, next row, row next to it and so on form the sequence 20, 19, 18, 17, ................
∴a2 – a1 = 19 – 20 = – 1
a3 – a2 = 18 – 19 = – 1
a4 – a3 = 17 – 18 = – 1
i.e., ak+1 – ak is the same everytime.
So, the above sequence forms an AP.
Here, a = 20 d = – 1 Sn = 200
We know that
⇒ 200 = n/2 [2 (20) + (n – 1) (– 1)] ⇒ 200 = n/2 [40 – n + 1]
⇒ 200 = n/2 [41 – n] ⇒ 400 = n [41 – n]⇒ n[41 – n] = 400
⇒ 41n – n2 = 400 ⇒ n2 – 41n + 400 = 0
⇒ n2 – 25n – 16n + 400 = 0
⇒ n (n – 25n) – 16 (n – 25) = 0 ⇒ (n – 25) (n – 16) = 0
⇒ n – 25 = 0 or n – 16 = 0
⇒ n = 25 or n = 16 ⇒ n = 25, 16
Hence, the number of rows is either 25 or 16.
Now, number of logs in row
= Number of logs in 25th row
= a25
= a + (25 – 1)d [∵ an = a + (n – 1)d]
= a + 24d = 20 + 24 (– 1)
⇒ 20 – 24 = – 4
Which is not possible.
Therefore, n = 16 and Number of log in top row
= Number of logs in 16th row
= a16 = a + (16 – 1) d [∵ an = a + (n – 1)d]
= a + 15 d = 20 + 15 (– 1)
= 20 – 15 = 5
Hence, the 200 logs are placed in 16 rows and there are 5 logs in the top row.
COMPETITION WINDOW
SUM OF n TERMS OF A G.P.
If Sn is the sum of first n terms of the G.P. a, ar, ar2,...,
i.e., Sn = a + ar + ar2 + .... + arn–1, then
r ≠ 1
Also, , r ≠ 1, where l is the last term i.e, the nth term.
For r = 1, Sn = na.
SUM OF AN INFINITE G.P.
The sum of an infinite G.P. with first term a and common ratio r, where – 1 < r < 1, is
If r ≥ 1, then the sum of an infinite G.P. tends to infinity.
SUM OF n TERMS OF A H.P.
There is no specific formula to find the sum of n terms of H.P. To solve the questions of this progression, first of all convert it in A.P. then use the properties of A.P.
TRY OUT THE FOLLOWING
1. Find the sum of seven terms of the G.P. 3, 6, 12, ...
2. Find the sum to 7 terms of the sequence
3. Find the sum of the series 2 + 6 + 18 +... + 4374. 4. How many terms of the sequence 1,
4, 16, 64,.... will make the sum 5461?
5. Find the sum to infinity of the G.P.
6. The first term of a G.P. is 2 and the sum of infinity is 6. Find the common ratio.
7. If each term of an infinite G.P. is twice the sum of the terms following it, then find the common ratio of the G.P.
ANSWERS
1. 381
2.
3. 6560
4. – 1
PROPERTI ES OF ARITHMETICAL PROGRESSIONS
1. If a constant is added to or subtracted from each term of an A.P., then the resulting sequence is also an A.P. with the same common difference.
2. If each term of a given A.P. is multiplied or divided by a non-zero constant K, then the resulting sequence is also an A.P. with common difference Kd or d/K, where d is the common difference of the given A.P.
3. In a finite A.P., the sum of the terms equidistant from the beginning and end is always same and is equal to the sum of first and last term.
4. Three numbers a,b,c are in A.P. iff 2b = a + c.
5. A sequence is an A.P. iff it's nth term is a linear expression in n i.e., an = An + B, where A, B are constants.
In such a case, the coefficient of n is the common difference of the A.P.
6. A sequence is an A.P. iff the sum of it's first n terms is of the form An2 + Bn, where A,B are constants independent of n. In such a case, the common difference is 2 A. 7. If the terms of an A.P. are chosen at regular intervals, then they form an A.P.
COMPETITION WINDOW
ARITHMETIC MEANS
1. If three numbers a,b,c are in A.P. then b is called the arithmetic mean (A.M.) between a and c.
2. The arithmetic mean between two numbers a and b is
3. A1, A2,....,An are said to be n A.M.s between two numbers a and b. iff a, A1, A2,....,An, b are in A.P.
Let d be the common difference of the A.P.
Clearly, b = (n + 2)th term of the A.P.
⇒ b = a + (n + 1) d
⇒ d =
Hence, A1 = a + d = a + , A2 = a + 2d = a +
...........................................................................
...........................................................................
An = a + nd = a +
4. The sum of n A.M.'s between two numbers a and b is n times the single A.M. between then i.e.,
GEOMETRIC MEANS
1. If three non-zero numbers a,b,c are in G.P., then b is called the geometric mean (G.M.) between a and b
2. The geometric mean between two positive numbers a and b is
HARMONIC MEAN
1. If three non-zero numbers a,b,c are in H.P., then b is called the harmonic mean (H.M.) between a and b.
2. The harmonic mean between numbers a and b is
Re ma rk : If A,G,H denote respectively, the A.M., the G.M. and the H.M. between two distinct positive numbers, then
(i) A,G,H are in G.P.
(ii) A > G > H ˜
5 videos|292 docs|59 tests
|
1. What is a G.P. (Geometric Progression)? |
2. How do you find the sum of the first N terms in a G.P.? |
3. How do you find the nth term in a G.P.? |
4. Can the common ratio in a G.P. be negative? |
5. What happens if the common ratio in a G.P. is greater than 1? |
5 videos|292 docs|59 tests
|
|
Explore Courses for Class 10 exam
|