SATHEE: Sequences and Series 1 Question 1

Sequences and Series 1 Question 1

1. If $a_{1}, a_{2}, a_{3}, \dots, a_{n}$ are in $A P$ and $a_{1} + a_{4} + a_{7} + \dots + a_{16}$ $= 114$ , then $a_{1} + a_{6} + a_{11} + a_{16}$ is equal to

(a) 64

(b) 76

(d) 38

(2019 Main, 10 April I)

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Answer:

Correct Answer: 1. (b)

Solution:

Key Idea Use $n$ th term of an AP i.e. $a_{n} = a + (n - 1) d$ , simplify the given equation and use result.

Given $A P$ is $a_{1}, a_{2}, a_{3}, \dots, a_{n}$

Let the above AP has common difference ’ $d$ ‘, then $a_{1} + a_{4} + a_{7} + \dots + a_{16}$

$= a_{1} + (a_{1} + 3 d) + (a_{1} + 6 d) + \dots + (a_{1} + 15 d)$

$= 6 a_{1} + (3 + 6 + 9 + 12 + 15) d$

$\begin{aligned} ∴ 6 a_{1} + 45 d = 114 \\ \Rightarrow 2 a_{1} + 15 d = 38 \end{aligned}$

(given)

Now, $a_{1} + a_{6} + a_{11} + a_{16}$

$= a_{1} + (a_{1} + 5 d) + (a_{1} + 10 d) + (a_{1} + 15 d)$

$= 4 a_{1} + 30 d = 2 (2 a_{1} + 15 d)$

$= 2 \times 38 = 76$

[from Eq. (i)]

Sequences and Series 1 Question 1

1. If $a_{1}, a_{2}, a_{3}, \dots, a_{n}$ are in $A P$ and $a_{1} + a_{4} + a_{7} + \dots + a_{16}$ $= 114$ , then $a_{1} + a_{6} + a_{11} + a_{16}$ is equal to

Answer:

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Sequences and Series 1 Question 1

1. If a1,a2,a3,…,an are in AP and a1+a4+a7+…+a16 =114, then a1+a6+a11+a16 is equal to

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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1. If $a_{1}, a_{2}, a_{3}, \dots, a_{n}$ are in $A P$ and $a_{1} + a_{4} + a_{7} + \dots + a_{16}$ $= 114$ , then $a_{1} + a_{6} + a_{11} + a_{16}$ is equal to