Question

How many terms of the A.P. 1,4,7,10... are needed to get the sum of 715?

Answer 1

Step-by-step explanation:

a(n)=715

d=7-4=3

using a(n)=a+(n-1)d

715=1+(n-1)3

714/3=(n-1)

238=(n-1)

n=239

hipe this helps you

Answer 2

Step-by-step explanation:

Here,

a=1

d= 4-1 =3

Sn=715

To find: 'n'

Solution:

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

715=n/2 [2×1 + (n-1) ×3]

715=n/2 [2 + 3n-3]

715=n +3n -3

715+3 =4n

718=4n

n= 718/4

n=179.5

n~=178

Ans: Therefore, 178 terms are needed to get the sum 715