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NDA 2023
Exam
NDA
Year
2023
Subject
Mathematics
Chapter
Sequences and Series

NDA 2023 · Mathematics · Sequences and Series

Let a_1, a_2, a_3 be in AP such that a_1+a_5+a_ 10 +a_ 15 +a_ 20...

Let a1,a2,a3…a_1, a_2, a_3 \ldots be in AP such that a1+a5+a10+a15+a20+a25+a30+a34=300a_1+a_5+a_{10}+a_{15}+a_{20}+a_{25}+a_{30}+a_{34}=300 What is ∑n=134an\sum_{n=1}^{34} a_n equal to?

Options

  1. A900
  2. B1025
  3. C1200
  4. D1275

Correct answer

D. 1275

Step-by-step solution

Given, a1+a5+a10+a15+a20+a25+a30+a34=300=a1+a1+4d+a1+9d+a1+14d+a1+19d+a1+24d+a1+29d+a1+33d=300=8a1+132d=300=(2a1+33d)=3004...(i)\begin{aligned} & a_1+a_5+a_{10}+a_{15}+a_{20}+a_{25}+a_{30}+a_{34}=300 \\ & =a_1+a_1+4 d+a_1+9 d+a_1+14 d+a_1+19 d \\ & +a_1+24 d+a_1+29 d+a_1+33 d=300 \\ & =8 a_1+132 d=300 \\ & =\left(2 a_1+33 d\right)=\frac{300}{4}...(i) \end{aligned} Since, ∑n=134an=a1+a2+………+a34=342(2a1+33d)=17×300=1275\begin{aligned} & \text{ Since, } \sum_{n=1}^{34} a_n=a_1+a_2+\ldots \ldots \ldots+a_{34} \\ & =\frac{34}{2}\left(2 a_1+33 d\right) \\ & =17 \times \frac{300}{}=1275 \end{aligned}
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