MHT CET · Maths · Sequences and Series
If for an Arithmetic progression, 9 times nineth term is equal to 13 times thirteenth term, then value of twenty second term is
- A 0
- B 2
- C 4
- D 5
Answer & Solution
Correct Answer
(A) 0
Step-by-step Solution
Detailed explanation
Let the first term and common difference of given \(\mathrm{AP}\) be a and d, respectively.
Given
\(\begin{array}{l}
9 . a_{9}=13 . a_{13} \\
\Rightarrow 9[a+(9-1) d]=13[a+(13-1) d] \\
\Rightarrow 9[a+8 d]=13[a+12 d] \\
\Rightarrow 9 a+72 d=13 a+156 d \\
\Rightarrow 4 a+84 d=0 \\
\Rightarrow 4[a+21 d]=0 \\
\Rightarrow 4[a+(22-1) d=0 \\
\Rightarrow a+(22-1) d=0 \\
\Rightarrow a_{22}=0
\end{array}\)
Given
\(\begin{array}{l}
9 . a_{9}=13 . a_{13} \\
\Rightarrow 9[a+(9-1) d]=13[a+(13-1) d] \\
\Rightarrow 9[a+8 d]=13[a+12 d] \\
\Rightarrow 9 a+72 d=13 a+156 d \\
\Rightarrow 4 a+84 d=0 \\
\Rightarrow 4[a+21 d]=0 \\
\Rightarrow 4[a+(22-1) d=0 \\
\Rightarrow a+(22-1) d=0 \\
\Rightarrow a_{22}=0
\end{array}\)
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