Question

5. Find the sum of 51 terms of A.P. in which II
and III rd terms are 14 and 18 respectively.​

Answer 1

Step-by-step explanation:

t=a+(n-1)d

so,

14=a+d

18=a+2d

solving simultaneously,

we get,

d=4

a=10

now , sun of n terms of A.P

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

substituting values ,

Sn=5610

Answer 2

Answer:

S₅₁=5610

Step-by-step explanation:

Given:-

a₂=14 and a₃=18

d=a₃-a₂=18-14

d=4

To Find:-

S₅₁

Formula Used:-

Sₙ=$\frac{n}{2}$[2a+(n-1)(d)]

Solution:-

a₃=a+2d

18=a+2(4)

a=18-8

a=10

Sₙ=n/2[2a+(n-1)d]

S₅₁=51/2[2(10)+(51-1)(4)]

S₅₁=51/2[20+(50)(4)]

S₅₁=51/2[20+200]

S₅₁=$\frac{51}{2}$×220

S₅₁=51×110

S₅₁=5610