Sum of Series in an Arithmetic Progression

Introduction

An arithmetic progression or arithmetic sequence (AP) is a sequence of numbers such that the difference from any succeeding term to its preceding term remains constant throughout the sequence. The constant difference is called common difference of that arithmetic progression. For instance, the sequence 5, 7, 9, 11, 13, 15, . . . is an arithmetic progression with a common difference of 2.

Verse

इष्टं व्येकं दलितं सपूर्वमुत्तरगुणं समुखं मध्यम् ।

इष्टगुणितमिष्टधनं त्वथवाद्यन्तं पदार्धहतम् ॥

Translation

Diminish the given number of terms by one, then divide by two, then increase by the number of the preceding terms (if any), then multiply by the common difference, and then increase by the first term of the (whole) series: the result is the arithmetic mean (of the given number of terms). This multiplied by the given number of terms is the sum of the given terms. Alternatively, multiply the sum of the first and last terms (of the series or partial series which is to be summed up) by half the number of terms.

Let an arithmetic series be $a+(a+d)+(a+2d)+..........$

Here a = First term; d = Common difference; n = no. of terms; p = no. of previous terms

As per the above rule

$The\ arithmetic\ mean\ of\ the\ n\ terms=M=(a+pd)+(a+(p+1)d)+......(a+(p+n-1)d)$

$=a+\left({\frac {n-1}{2}}+p\right)d$

$The\ sum\ of\ the\ n\ terms=no.\ of\ terms\times Arithmetic\ mean$

$=n\times M$

$=n\left[a+\left({\frac {n-1}{2}}+p\right)d\right]$

$The\ nth\ term=a+(n-1+p)\times d$

In particular when p = no. of previous terms = 0

$The\ arithmetic\ mean\ of\ the\ n\ terms=M=a+(a+d)+......(a+(n-1)d)$

$=a+\left({\frac {n-1}{2}}\right)d$

$The\ sum\ of\ the\ n\ terms=S=no.\ of\ terms\times Arithmetic\ mean$

$=n\times M$

$=n\left[a+\left({\frac {n-1}{2}}\right)d\right]$

$The\ nth\ term=a+(n-1)\times d$

Alternatively, the sum of n terms of an arithmetic series with A as the first term and L as the last term

$The\ sum\ of\ the\ n\ terms=S={\frac {n}{2}}(A+L)$

Examples

Example 1

For the series 1, 5, 9, 13, 17, 21, 25, 29, 33, 37 find the first term, common difference, no. of terms, last term, sum of series.

First term = A	1
Common Difference = d	5-1 =4
No. of terms = n	10
Last term = L	37
Sum of series = S	$={\frac {n}{2}}(A+L)={\frac {10}{2}}(1+37)=5(38)=190$

Example 2

For a certain series first term is 2, common difference is 3, no. of terms 5. Find the mean and sum of series.

Here first term a = 2, common difference d = 3, no. of terms n = 5.

Mean M $=a+\left({\frac {n-1}{2}}\right)d=2+\left({\frac {5-1}{2}}\right)3=8$

$The\ sum\ of\ the\ n\ terms=no.\ of\ terms\times Arithmetic\ mean$

$=n\times M=5\times 8=40$

Example 3

For a certain series first term is 7, common difference is 11, no. of terms 25. Find the last term, penultimate term and 20^th term..

Here first term a = 7, common difference d = 11, no. of terms n = 25.

$The\ nth\ term(n=25)=a+(n-1)\times d=7+(25-1)\times 11=7+24\times 11=271$

Penultimate term = no. of terms - 1 = 25 - 1 = 24

$The\ nth\ term(n=24)=a+(n-1)\times d=7+(24-1)\times 11=7+23\times 11=260$

20th term :

$The\ nth\ term(n=20)=a+(n-1)\times d=7+(20-1)\times 11=7+19\times 11=216$

See Also

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