27.6k views
4 votes
The sum of terms of an A.P. is 136, the common difference 4, and the last term 31, find n​

User KhanS
by
8.4k points

2 Answers

10 votes

Answer:

Sn=2n{2a+(n−1)d}

a+(n−1)d=31⇒a=31−4(n−1)

∴136=2n{2[31−4(n−1)]+(n−1)d}

n[70−8n+4n−4]=272

n(66−4n)=272

4n2−66n+272=0

2n2−33n+136=0

2n(n−8)−17(n−8)=0

(2n−17)(n−8)=0

∴n=8

User McG
by
7.4k points
6 votes

Explanation:


\huge{\color{black}{\fbox{\color{pink}{\colorbox{pink}{\color{red}{Answer ✨☑️}}}}}}

The number of terms are 8. Step-by-step explanation: Given : The sum n terms of an A.P is 136 the common difference is 4 and last term is 31.

User Yuval Itzchakov
by
8.1k points

No related questions found

Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.