Question

How many terms of the AP 25,28,31,34,...... are needed to give the sum 1070?

Answer 1

I got the enough solution of it ...

After solving we will get n = 20

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Answer 2

Given :-

AP - 25, 28, 31, 34 ...

Here a (first term) = 25, d (common difference) = 28 - 25 = 3

Let the sum of n terms be 1070

➡ Sn = 1070

➡ n/2 [2a + (n - 1)d] = 1070

➡ n/2 [(2 × 25) + (n - 1)3] = 1070

➡ n/2 (50 + 3n - 3) = 1070

➡ n(47 + 3n) = 1070 × 2

➡ 47n + 3n² = 2140

➡ 3n² + 47n - 2140 = 0

By factorisation, we get

➡ 3n² + (107n - 60n) - 2140 = 0

➡ 3n² + 107n - 60n - 2140 = 0

➡ n(3n + 107) - 20(3n + 107) = 0

➡ (3n + 107) (n - 20)

➡ n = -107/3 or n = 20

Since n can't be negative.

Therefore n = 20

Hence, the sum of first 20 terms is 1070.