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Chapter 5 · Question 6

Find the sum of the first 15 terms of the AP 5,8,11,…5,8,11,\ldots.

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Answer

Direct Answer

Here a=5a=5, d=3d=3, n=15n=15. S15=152[2(5)+14(3)]=152(52)=390S_{15}=\frac{15}{2}[2(5)+14(3)]=\frac{15}{2}(52)=390.

Simple Explanation

Use the AP sum formula. The sum is 390390.

Exam-Ready Structure

For 5,8,11,…5,8,11,\ldots, a=5a=5 and d=3d=3. Then Sn=n2[2a+(n−1)d]S_n=\frac n2[2a+(n-1)d]. Thus S15=152[10+14×3]=152(52)=390S_{15}=\frac{15}{2}[10+14\times3]=\frac{15}{2}(52)=390.

Key Points

  • Here a=5a=5, d=3d=3, n=15n=15.

Related Questions

Q1

What is an arithmetic progression? Find the common difference of 45,43,41,39,…45,43,41,39,\ldots.

Q2

State the formula for the nnth term of an AP and use it to find the 20th term of 3,7,11,15,…3,7,11,15,\ldots.

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