Question

find the sum of series 62+60+58+....+40​

Answer 1

SOLUTION

TO DETERMINE

The sum of series 62+60+58+....+40

EVALUATION

Here the given series is

62+60+58+....+40

This is an arithmetic progression

First term = a = 62

Common Difference = d = 60 - 62 = - 2

Let number of terms = n

Then

$\sf{a + (n-1)d= 40}$

$\sf{ \implies \: 62 + (n-1) \times - 2= 40}$

$\sf{ \implies \: - 2(n-1) \times = - 22}$

$\sf{ \implies \: (n-1) = 11}$

$\sf{ \implies \: n = 12}$

Hence the required sum

$\displaystyle \sf{ = \frac{n}{2} \bigg [ \:2a + (n - 1)d \: \bigg]}$

$\displaystyle \sf{ = \frac{12}{2} \bigg [ \:2 \times 62 + (12 - 1) \times ( - 2) \: \bigg]}$

$\displaystyle \sf{ = 6 \bigg [124 - 22 \: \bigg]}$

$\displaystyle \sf{ = 6 \times 102}$

$\displaystyle \sf{ = 612}$

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Answer 2

Step-by-step explanation:

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