Page No 5.52
Ques.14. The first and the last terms of an A.P. are 17 and 350 respectively. If the common difference is 9, how many terms are there and what is their sum?
Ans. In the given problem, we have the first and the last term of an A.P. along with the common difference of the A.P. Here, we need to find the number of terms of the A.P. and the sum of all the terms.
Here,
The first term of the A.P (a) = 17
The last term of the A.P (l) = 350
The common difference of the A.P. = 9
Let the number of terms be n.
So, as we know that,
l = a + (n - 1)d
We get,
350 = 17 + (n - 1)9
350 = 17 + 9n - 9
350 = 8 + 9n
350 - 8 = 9n
Further solving this,
n = 342/9 = 38
Using the above values in the formula,
= (19)(367) = 6973
Therefore, the number of terms is n = 38 and the sum Sn = 6973.
Ques.15. The third term of an A.P. is 7 and the seventh term exceeds three times the third term by 2. Find the first term, the common difference and the sum of first 20 terms.
Ans. In the given problem, let us take the first term as a and the common difference as d.
Here, we are given that,
a3 = 7 ...(1)
a7 = 3a3 + 2 ...(2)
So, using (1) in (2), we get,
a7 = 3(7) + 2
= 21 + 2 = 23 ...(3)
Also, we know,
an = a + (n - 1)d
For the 3th term (n = 3),
a3 = a + (3 - 1)d
7 = a + 2d (Using 1)
a = 7 - 2d ...(4)
Similarly, for the 7th term (n = 7),
a7 = a + (7 - 1)d
24 = a + 6d (Using 3)
a = 24 - 6d ...(5)
Subtracting (4) from (5), we get,
a - a = (23 - 6d)-(7 - 2d)
0 = 23 - 6d - 7 + 2d
0 = 16 - 4d
4d = 16
d = 4
Now, to find a, we substitute the value of d in (4),
a = 7 - 2(4)
a = 7 - 8 = -1
So, for the given A.P d = 4 and a = -1
So, to find the sum of first 20 terms of this A.P., we use the following formula for the sum of n terms of an A.P.,
Sn = n/2[2a + (n - 1)d]
Where; a = first term for the given A.P.
d = common difference of the given A.P.
n = number of terms
So, using the formula for n = 20, we get,
S20 = 20/2[2(-1) + (20 - 1)(4)]
= (10)[-2 + (19)(4)]
= (10)[-2 + 76)
= (10)[74] = 740
Therefore, the sum of first 20 terms for the given A.P. is S20 = 740.
Ques.16. The first term of an A.P. is 2 and the last term is 50. The sum of all these terms is 442. Find the common difference.
Ans. In the given problem, we have the first and the last term of an A.P. along with the sum of all the terms of A.P. Here, we need to find the common difference of the A.P.
Here,
The first term of the A.P (a) = 2
The last term of the A.P (l) = 50
Sum of all the terms Sn = 442
Let the common difference of the A.P. be d.
So, let us first find the number of the terms (n) using the formula,
442 = (n/2)(2 + 50)
442 = (n/2)(52)
442 = (n)(26)
n = 442/26 = 17
Now, to find the common difference of the A.P. we use the following formula,
l = a + (n - 1)d
We get,
50 = 2 + (17 - 1)d
50 = 2 + 17d - d
50 = 2 + 16d
Further, solving for d,
d = 48/16 = 3
Therefore the common difference of the A.P d = 3.
Ques.17. If 12th term of an A.P. is −13 and the sum of the first four terms is 24, what is the sum of first 10 terms.
Ans. In the given problem, we need to find the sum of first 10 terms of an A.P. Let us take the first term a and the common difference as d
Here, we are given that,
a12 = -13
S4 = 24
Also, we know,
an = a + (n - 1)d
For the 12th term (n = 12),
a12 = a + (12 - 1)d
-13 = a + 11d
a = -13 - 11d ...(1)
So, as we know the formula for the sum of n terms of an A.P. is given by,
Sn = n/2[2a + (n - 1)d]
Where; a = first term for the given A.P.
d = common difference of the given A.P.
n = number of terms
So, using the formula for n = 4, we get,
S4 = 4/2[2(a) + (4 - 1)(d)]
24 = (2)[2a + (3)(d)]
24 = 4a + 6d
4a = 24 - 6d
a = 6 - (6/4)d ...(2)
Subtracting (1) from (2), we get,
0 = 6 - (6/4)d + 13 + 11d
0 = 19 + 11d - 6/4(d)
On further simplifying for d, we get,
d = -2
Now, to find a, we substitute the value of d in (1),
a = -13 - 11(-2)
a = -13 + 22
a = 9
Now, using the formula for the sum of n terms of an A.P. for n = 10, we get,
S10 = (10/2)[2(9) + (10 - 1)(-2)]
= (5)[18 + (9)(-2)]
= (5)(18 - 18)
= (5)(0) = 0
Therefore, the sum of first 10 terms for the given A.P. is S10 = 0.
Ques.18. Find the sum of n terms of the series ...
Ans. Let the given series be S = ...
= S1 − S2
S1 = 4[1 + 1 + 1 + ...]
a = 1, d = 0
S1 = 4 x n/2 [2 x 1 + (n - 1) x 0] (Sn = (n/2)(2a + (n - 1)d))
⇒ S1 = 4n
S2 = (1/n)[1 + 2 + 3 + ...]
a = 1, d = 2 − 1 = 1
Hence, the sum of n terms of the series is .
Ques.19. In an A.P., if the first term is 22, the common difference is −4 and the sum to n terms is 64, find n.
Ans. In the given problem, we need to find the number of terms of an A.P. Let us take the number of terms as n.
Here, we are given that,
a = 22
d = -4
Sn = 64
So, as we know the formula for the sum of n terms of an A.P. is given by,
Sn = n/2[2a + (n - 1)d]
Where; a = first term for the given A.P.
d = common difference of the given A.P.
n = number of terms
So, using the formula we get,
64(2) = n(48 - 4n)
128 = 48n - 4n2
Further rearranging the terms, we get a quadratic equation,
4n2 - 48n + 128 = 0
On taking 4 common, we get,
n2 - 12n + 32 = 0
Further, on solving the equation for n by splitting the middle term, we get,
n2 - 12n + 32 = 0
n2 - 8n - 4n + 32 = 0
n(n - 8)-4(n - 8) = 0
(n - 8)(n - 4) = 0
So, we get,
(n - 8) = 0
n = 8
Also,
(n - 4) = 0
n = 4
Therefore, n = 4 or 8.
Ques.20. In an A.P., if the 5th and 12th terms are 30 and 65 respectively, what is the sum of first 20 terms?
Ans. In the given problem, let us take the first term as a and the common difference d
Here, we are given that,
a5 = 30 ...(1)
a12 = 65 ...(2)
Also, we know,
an = a + (n - 1)d
For the 5th term (n = 5),
a5 = a + (5 - 1)d
30 = a + 4d (Using 1)
a = 30 - 4d ...(3)
Similarly, for the 12th term (n = 12),
a12 = a + (12 - 1)d
65 = a + 11d (Using 2)
a = 65 - 11d ...(4)
Subtracting (3) from (4), we get,
a - a = (65 - 11d) - (30 - 4d)
0 = 65 - 11d - 30 + 4d
0 = 35 - 7d
7d = 35
d = 5
Now, to find a, we substitute the value of d in (4),
a = 30 - 4(5)
a = 30 - 20 = 10
So, for the given A.P d = 5 and a = 10
So, to find the sum of first 20 terms of this A.P., we use the following formula for the sum of n terms of an A.P.,
Sn = n/2[2a + (n - 1)d]
Where; a = first term for the given A.P.
d = common difference of the given A.P.
n = number of terms
So, using the formula for n = 20, we get,
S20 = 20/2[2(10) + (20 - 1)(5)]
= (10)[20 + (19)(5)]
= (10)[20 + 95]
= (10)[115] = 1150
Therefore, the sum of first 20 terms for the given A.P. is S20 = 1150.
Ques.21. Find the sum of first 51 terms of an A.P. whose second and third terms are 14 and 18 respectively.
Ans. In the given problem, let us take the first term as a and the common difference as d.
Here, we are given that,
a2 = 14 ...(1)
a3 = 18 ...(2)
Also, we know,
an = a + (n - 1)d
For the 2nd term (n = 2),
a2 = a + (2 - 1)d
14 = a + d (Using 1)
a = 14 - d ...(3)
Similarly, for the 3rd term (n = 3),
a3 = a + (3 - 1)d
18 = a + 2d (Using 2)
a = 18 - 2d ...(4)
Subtracting (3) from (4), we get,
a - a = (18 - 2d)-(14 - d)
0 = 18 - 2d - 14 + d
0 = 4 - d
d = 4
Now, to find a, we substitute the value of d in (4),
a = 14 - 4 = 10
So, for the given A.P d = 4 and a = 10
So, to find the sum of first 51 terms of this A.P., we use the following formula for the sum of n terms of an A.P.,
Sn = n/2[2a + (n - 1)d]
Where; a = first term for the given A.P.
d = common difference of the given A.P.
n = number of terms
So, using the formula for n = 51, we get,
S51 = 51/2 [2(10) + (51 - 1)(4)]
= 51/2 [20 + (50)(4)]
= 51/2 [20 + 200]
= 51/2 [220]
= 51(110) = 5610
Therefore, the sum of first 51 terms for the given A.P. is S51 = 5610
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