NIMCET 2025 Paper

Section: Mathematics

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Question : 12 of 120

Marks: +1, -0

Suppose

t_1, t_2, t_3, . . ., t_{55}

are in A.P. such that

∑↙{l=0}↖{18} t_{3 l+1}=1197 \;\; \;\text " and "\; \;\; t_7+3 t_{22}=174

If

∑↙{l=1}↖{9} t_l^2=947 b

then the value of

b

is:

5
3
2
1

Solution:

Let

t_n=a+(n-1) d

.
1. Use

∑↙{l=0}↖{18} t_{3 l+1}=1197

.
The terms are

t_1, t_4, t_7, . . ., t_{55}: 19

terms, A.P. with first term

t_1=a

, common difference

3 d

.

∑=\;{19}/{2}[2 a+(19-1) ⋅ 3 d]=1197

\;{19}/{2}(2 a+54 d)=1197 ⇒ 2 a+54 d=126 ⇒ a+27 d=63

2. Use

t_7+3 t_{22}=174

.

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

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

a+6 d+3 a+63 d=174 ⇒ 4 a+69 d=174

From (1):

a=63-27 d

. Substitute in (2):

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

252-108 d+69 d=174 ⇒ 252-39 d=174 ⇒ d=2

a=63-27 ⋅ 2=9

So

t_n=9+(n-1) ⋅ 2

First 9 terms:

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

.
Sum of squares:

∑↙{l=1}↖{9} t_l^2=9^2+11^2+⋯+25^2=2841

Given:

2841=947 b ⇒ b=\;{2841}/{947}=3

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