Find $S_{15}$ of an A.P: $4, 8, 12, 16, . . . . . .$

Open in App

Solution

$a = 4$ ,
$d = t_{2} = t_{1} = 8 - 4 = 4$ , where $t_{2}, t_{1}$ are the 1st and 2nd term of the A.P.
$S_{15} = \frac{n}{2} [2 a + (n - 1) d]$
$= \frac{15}{2} [2 \times 4 + (15 - 1) \times 4]$
$= \frac{15}{2} \times [8 + 56]$
$= \frac{15}{2} \times 64$
$= 480$

Suggest Corrections

Similar questions

S_{15}

4, 8, 12, . . .

S_{15}

2, 5, 8, . . . . ?

S_{15}

a = 10, d = 13

a_{1} = 1.5, a_{20} = 58.5

S_{15}

S_{15}

n t h

a_{n}

= 3 + 4 n

Join BYJU'S Learning Program