Question

If for an ap S31=186 then t15=?

Answer 1

I think question is .....

if for an AP S31 = 186 then, t16 = ?

it is given that,

sum of 31 terms in arithmetic progression , $S_{31}=186$

we know, $S_n=\frac{n}{2}[2a+(n-1)d]$

here n = 31, $S_{31}=186$

so, 186 = 31/2 [ 2a + (31 - 1)d ]

or, 186 × 2/31 = 2a + 30d

or, 12 = 2a + 30d

or, a + 15d = 6.........(1)

now, using formula, Tn = a + (n - 1)d

16th term , $T_{16}$ = a + (16 - 1)d = a + 15d

from equation (1), $T_{16}$ = 6