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Sequences and Series question

2019 · 10 Apr · Shift 1 · Q33

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Sequences and Series
2019 · 10 Apr · Shift 1 · Q33

Sequences and Series question

2019 · 10 Apr · Shift 1 · Q33

JEE MainMathematicsSequences and SeriesMCQ+4 / −1

If a1, a2, a3, ............... an are in A.P. and a1 + a4 + a7 + ........... + a16 = 114, then a1 + a6 + a11 + a16 is equal to :

A
38
B
98
C
76
D
64

View written solutionFree

Correct answer: C

Let the A.P. be $a_n = a + (n-1)d$ where $a_1=a$ .
First, use the given sum: $a_1+a_4+a_7+a_{10}+a_{13}+a_{16}=114$

Now write each term: [ a_1=a, \quad a_4=a+3d, \quad a_7=a+6d, \quad a_{10}=a+9d, \quad a_{13}=a+12d, \quad a_{16}=a+15d ]

So, [ a_1+a_4+a_7+a_{10}+a_{13}+a_{16} =6a+(3+6+9+12+15)d =6a+45d ]

Hence, $6a+45d=114$
We need to find: $a_1+a_6+a_{11}+a_{16}$

Write these terms: [ a_1=a, \quad a_6=a+5d, \quad a_{11}=a+10d, \quad a_{16}=a+15d ]

Therefore, [ a_1+a_6+a_{11}+a_{16} =4a+(5+10+15)d =4a+30d ]
From $6a+45d=114$ divide by $\frac{3}{2}$ or factor directly: $6a+45d=3(2a+15d)=114$ so $2a+15d=38$

Then, $4a+30d=2(2a+15d)=2\times 38=76$
Therefore, $a_1+a_6+a_{11}+a_{16}=76$
Checking options:
- A: $38$
- B: $98$
- C: $76$ ✅
- D: $64$

So the correct option is C.

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