Question

In the A.P. 7, 14, 21, ... How many terms are to be considered for getting sum 5740.

Answer 1

In the given A.P., we have:
a = 7 and d = t₂ – t₁ = 14 – 7 = 7
Let the sum of n terms of the given A.P. be S_n = 5740.
We know:
Sum of n terms of an A.P., $S_n=\frac{n}{2}\left[2a+\left(n-1\right)d\right]$
Thus, we have:

$5740=\frac{n}{2}\left[2\times 7+\left(n-1\right)\times 7\right]$

$\Rightarrow$ 2 × 5740 = n[14 + 7n – 7] = n[7n + 7]
$\Rightarrow$ 11480 = 7n² + 7n
$\Rightarrow$ 7n² + 7n – 11480 = 0
$\Rightarrow$ 7[n² + n – 1640] = 0
$\Rightarrow$ n² + n – 1640 = 0
On splitting the middle term, we get:
$\Rightarrow$ n² + 41n – 40n – 1640 = 0
$\Rightarrow$ n(n + 41) – 40(n + 41) = 0
$\Rightarrow$ (n + 41)(n – 40) = 0
$\Rightarrow$ n + 41 = 0 or n – 40 = 0
$\Rightarrow$ n = –41 or n = 40

Because n is the number of terms that cannot be negative, n is 40.