Question

How many terms of the A.P. 7, 14, 21, ... are
needed to give the sum 5740?​

Answer 1

Given ,

The AP is 7 , 14 , 21, ...

First term (a) = 7

Common difference (d) = 7

Sum (Sn) = 5740

We know that , the sum of first n terms of an AP is given by

$\boxed{ \tt{S_{n} = \frac{n}{2} \{2a + (n - 1)d \}}}$

Thus ,

5740 = n/2 × {2 × 7 + (n - 1)7}

11480 = n × {14 + (n -1) 7}

11480 = n × {14 + 7n - 7}

11480 = 14n + 7n² - 7n

11480 = 7n² + 7n

7n² + 7n - 11480 = 0

n² + n - 1640 = 0

n² + 41n - 40n - 1640 = 0

n(n + 41) - 40(n + 41) = 0

(n - 40)(n + 41) = 0

n = 40 or n = -41

Since , n ≠ negative

Therefore , the first 40 terms of given AP is 5740

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