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Sequences and Series question

2019 · 9 Apr · Shift 2 · Q26

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2019 · 9 Apr · Shift 2 · Q26

Sequences and Series question

2019 · 9 Apr · Shift 2 · Q26

JEE MainMathematicsSequences and SeriesMCQ+4 / −1

If the sum and product of the first three term in an A.P. are 33 and 1155, respectively, then a value of its 11th term is :-

A
–25
B
–36
C
25
D
–35

View written solutionFree

Correct answer: A

Let the first three terms of the A.P. be $a-d,\ a,\ a+d$ where $a$ is the middle term and $d$ is the common difference.
Use the sum condition The sum of the first three terms is given as $33$ : $(a-d)+a+(a+d)=33$ $3a=33$ $a=11$
Use the product condition The product of the first three terms is $1155$ : $(a-d)\cdot a \cdot (a+d)=1155$ Substitute $a=11$ : $(11-d)(11)(11+d)=1155$ $11(121-d^2)=1155$ $121-d^2=105$ $d^2=16$ $d=\pm 4$
Find the 11th term If the first term is $a-d=11-d$ , then the 11th term is: $T_{11}=a_1+10d=(11-d)+10d=11+9d$

Now check both possible values of $d$ :
- If $d=4$ , $T_{11}=11+9(4)=11+36=47$
- If $d=-4$ , $T_{11}=11+9(-4)=11-36=-25$
Match with the options Among the given options, only $-25$ appears.

Therefore, the required value of the 11th term is: $\boxed{-25}$

So, the correct option is A.

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