Question

1. Tf
3+ 5 + 7 + ... + n
-= 7, then the value of
5+8+11+ ... + 10 terms
n is
(2) 35
(6) 36
(c) 37
(d) 40​

Answer 1

Step-by-step explanation:

$\frac{n }{2}(2a +( n - 1)d)$

So,

$\frac{ \frac{n}{2}(6) + (n - 1)2) }{ \frac{10}{2}(10 + 27) } = 7$

$\frac{n(6 + 2n - 2)}{2} = 7 \times 5 \times 37$

$n(4 + 2n) = 7 \times 5 \times 37 \times 2$

$4n + 2 {n }^{2} - 2590 = 0$

$2n + {n}^{2} - 1295 = 0$

${n}^{2} + 37n - 35n - 1295 = 0$

$(n - 35)(n + 37) = 0$

'n' should not be negative

So, n=35

${n}^{2} + 2n - 1295 = 0$