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NIMCET 2025 Paper

Section: Mathematics

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Question : 12 of 120

Marks: +1, -0

Suppose

t_1, t_2, t_3, \ldots, t_{55}

are in A.P. such that

\sum\limits_{l=0}^{18} t_{3l+1}=1197 \qquad \text{ and } \qquad t_7+3t_{22}=174

\sum\limits_{l=1}^{9} t_l^2=947 b

then the value of

b

5
3
2
1

Solution:

Let

t_n=a+(n-1)d

.

1. Use

\sum\limits_{l=0}^{18} t_{3l+1}=1197

.

The terms are

t_1, t_4, t_7, \ldots, t_{55}: 19

terms, A.P. with first term

t_1=a

, common difference

3d

.

\sum=\frac{19}{2}[2a+(19-1)\cdot 3d]=1197

\frac{19}{2}(2a+54d)=1197 \Rightarrow 2a+54d=126 \Rightarrow a+27d=63

2. Use

t_7+3t_{22}=174

.

t_7=a+6d, \qquad t_{22}=a+21d

(a+6d)+3(a+21d)=174

a+6d+3a+63d=174 \Rightarrow 4a+69d=174

From (1):

a=63-27d

. Substitute in (2):

4(63-27d)+69d=174

252-108d+69d=174 \Rightarrow 252-39d=174 \Rightarrow d=2

a=63-27\cdot 2=9

So

t_n=9+(n-1)\cdot 2

First 9 terms:

9,11,13,15,17,19,21,23,25

.

Sum of squares:

\sum\limits_{l=1}^{9} t_l^2=9^2+11^2+\cdots+25^2=2841

Given:

2841=947b \Rightarrow b=\frac{2841}{947}=3

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