The sum of first third & seventeenth terms of an A.P.is 216 .Find the sum of the first 13 terms Of an A.P. About the author Maria
Answer: sum of the first 13 terms of the given AP = 936 Step-by-step explanation: we know, nth term of an AP , tₙ = a+(n-1)d => 1st term = a ; 3rd term =a+2d ; 17th term = a+16d; Given that, a+(a+2d)+(a+16d)=216 3a+18d = 216 3(a+6d) = 216 a+6d = 216/3 a+6d = 72 ——— (1) Sum of “n” terms of an AP, Sₙ = (n/2) [2a+(n-1)d] => S₁₃ = (13/2)[2a+12d] = (13/2)[2(a+6d)] = 13(a+6d) = 13*72 [ by (1) ] = 936 Reply
Answer:
sum of the first 13 terms of the given AP = 936
Step-by-step explanation:
we know, nth term of an AP , tₙ = a+(n-1)d
=> 1st term = a ;
3rd term =a+2d ;
17th term = a+16d;
Given that,
a+(a+2d)+(a+16d)=216
3a+18d = 216
3(a+6d) = 216
a+6d = 216/3
a+6d = 72 ——— (1)
Sum of “n” terms of an AP, Sₙ = (n/2) [2a+(n-1)d]
=> S₁₃ = (13/2)[2a+12d]
= (13/2)[2(a+6d)]
= 13(a+6d)
= 13*72 [ by (1) ]
= 936