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# Arithmetic Progression - Examples (with Solutions), Algebra, Quantitative Aptitude GMAT Notes | EduRev

## SSC : Arithmetic Progression - Examples (with Solutions), Algebra, Quantitative Aptitude GMAT Notes | EduRev

The document Arithmetic Progression - Examples (with Solutions), Algebra, Quantitative Aptitude GMAT Notes | EduRev is a part of the SSC Course UPSC Prelims Paper 2 CSAT - Quant, Verbal & Decision Making.
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Sequence & Series

A set of numbers whose domain is a real number is called a SEQUENCE and sum of the sequence is called a
SERIES. If is a sequence, then the expression is
a series.
Those sequences whose terms follow certain patterns are called progressions.

For example Also if f (n) = n2 is a sequence, then f (10) = 102 = 100 and so on.

The nth term of a sequence is usually denoted by Tn
Thus T1 = first term, T2 = second term, T10 = tenth term and so on.

There are three different progressions

1. Arithmetic Progression (A.P)
2. Geometric Progression (G.P)
3. Harmonic Progression (H.P)

Arithmetic Progression
It is a series in which any two consecutive terms have common difference and next term can be derived by
adding that common difference in the previous term.
Therefore Tn+1 - Tn = constant and called common difference (d) for all n ∈ N.

Examples:

1. 1, 4, 7, 10, ……. is an A. P. whose first term is 1 and the common difference is
d = (4 - 1) = (7 - 4) = (10 - 7) = 3.

2. 11, 7, 3, - 1 …… is an A. P. whose first term is 11 and the common difference
d = 7 - 11 = 3 - 7, = - 1 - 3 = - 4.

If in an A. P.

a = first term,
d = common difference = Tn - Tn-1
Tn = nth term Then a, a + d, a + 2d, a + 3d,... are in A.P.
nth term of an A.P.

The nth term of an A.P is given by the formula Note: If the last term of the A.P. consisting of n terms be l , then
l = a + (n - 1) d

Sum of n terms of an A.P
The sum of first n terms of an AP is usually denoted by Sn and is given by the following formula:  Where ‘l ’ is the last term of the series.

Ex.1 Find the series whose nth term is . Is it an A. P. series? If yes, find 101st term.

Sol. Putting 1, 2, 3, 4…. We get T1, T2, T3, T4………….. As the common differences are equal
∴The series is an A.P. Ex.2 Find 8th, 12th and 16th terms of the series; - 6, - 2, 2, 6, 10, 14, 18…

Sol.  Properties of an AP
I. If each term of an AP is increased, decreased, multiplied or divided by the
same non-zero number, then the resulting sequence is also an AP.

For example: For A.P. 3, 5, 7, 9, 11…  II. In an AP, the sum of terms equidistant from the beginning and end is always same and equal to the sum
of first and last terms as shown in example below.

III. Three numbers in AP are taken as a - d, a, a + d.
For 4 numbers in AP are taken as a - 3d, a - d, a + d, a + 3d.
For 5 numbers in AP are taken as a - 2d, a - d, a, a + d, a + 2d.

IV. Three numbers a, b, c are in A.P. if and only if
2b = a + c. and b is called Arithmetic mean of a & c

Ex.3 The sum of three numbers in A.P. is - 3, and their product is 8. Find the numbers.

Sol. Let the numbers be (a - d), a, (a + d). Then,
Sum = - 3 ⇒ (a - d) + a + (a + d) = - 3 ⇒ 3a = - 3 ⇒ a = - 1
Product = 8
⇒ (a - d) (a) (a + d) = 8
⇒ a (a2 - d2) = 8
⇒ (-1) (1 - d2) = 8
⇒ d2 = 9
⇒ d = ± 3
If d = 3, the numbers are - 4, - 1, 2. If d = - 3, the numbers are 2, - 1, - 4.
Thus, the numbers are - 4, - 1, 2 or 2, - 1, - 4.

Ex.4 A student purchases a pen for Rs. 100. At the end of 8 years, it is valued at Rs. 20. Assuming
that the yearly depreciation is constant. Find the annual depreciation.

Sol. Original cost of pen = Rs. 100
Let D be the annual depreciation.
∴ Price after one year = 100 - D = T1 = a (say)
∴ Price after eight years = T8 = a + 7 (- D) = a - 7D
= 100 - D - 7D = 100 - 8D
By the given condition 100 - 8D = 20
8D = 80

∴D = 10.
Hence annual depreciation = Rs. 10.

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## UPSC Prelims Paper 2 CSAT - Quant, Verbal & Decision Making

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