Question

Find the sum of the last ten terms of the A.P: 8, 10, 12, …., 126.

Answer 1

Hint: Here we have a series of terms in A.P. To find the sum of last ten terms, first we need to reverse the given A.P. According to the reverse A.P we have the first term a = 126, common difference (d) = -2, and number of terms (n) = 10. We will put these values in the formula of sum of AP. In this way we can find the sum of the last ten terms of the A.P.

Complete step-by-step answer:
Given: A.P: 8, 10, 12, …, 126.
Reversing all terms of given AP, we get another AP i.e.
 A.P: 126, 124, 122, …., 12, 10, 8
Here from reverse A.P we have;
a = 126, d = 124 – 126 = −2, n = 10
We have formula of sum of n terms of an AP
\[{S_n} = \dfrac{n}{2}\left[ {2a + (n - 1)d} \right]\]
To find the sum of 10 terms, we put values of a and d and n = 10.
${S_{10}} = \dfrac{{10}}{2}\left[ {2 \times 126 + (10 - 1)( - 2)} \right]$
${S_{10}} = 5\left[ {252 - 18} \right] = 5 \times 234 = 1170$
Hence, the sum of the last ten terms of the A.P: 8, 10, 12, …., 126 is 1170.

Note: An arithmetic progression is a sequence and series of numbers, in which the difference of any two numbers is a constant value. For example – Series of odd numbers – 1, 3, 5, 7, … is in A.P, which has a first term that is a = 1, common difference between two successive terms is equal to 2. In Arithmetic progression every pair of consecutive terms, the second number is obtained by adding a fixed number to the first one. That fixed number is known as common difference of the A.P. Sum of the n terms of the arithmetic progression is \[{S_n} = \dfrac{n}{2}\left[ {2a + (n - 1)d} \right]\].