Question

if the sum of first 6 terms is 96 and sum of first 10 terms is 240 . find the sum of n term of AP​

Answer 1

Step-by-step explanation:

S=n/2[2a+(n-1)d]

sum of first 6 terms = 96

S=6/2[2a+(6-1)d]=96

3[2a+5d]=96

2a+5d=32- - -1

sum of first 10 terms = 240

S=10/2[2a+(10-1)d]=240

5[2a+9d]=240

2a+9d=48- - -2

on subtracting 1&2 and then solving we get,

4d=16

d=4

now putting value of d in 1 we get,

2a+5(4)=32

2a+20=32

2a=12

a=6

now,

AP formed will be: 6,10,14.....

sum of n term of AP:

n/2[2a+(n-1)d]

n/2[2(6)+(n-1)(4)]

n/2[12+4n-4]

n/2[4n+8]

n/2[(4)(n+2)]

2n(n+2)

2n²+4n