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If $5^{\text {th }}$ term of a H.P. is $\frac{1}{45}$ and $11^{\text{th }}$ term is $\frac{1}{69}$, then its $16^{\text {th }}$ term will be
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The correct answer is:
$1 / 89$
Here $5^{\text {th }}$ term of the corresponding
A.P. $=a+4 d=45$\ldots(i)
and $11^{\text {th }}$ term of the corresponding
A.P. $=a+10 d=69$\ldots(ii)
From (i) and (ii), we get$a=29, d=4$
Therefore $16^{\text {th }}$ term of the corresponding A.P.
$=a+15 d=29+15 \times 4=89$
Hence $16^{\text {th }}$ term of the H.P. is $\frac{1}{89}$
A.P. $=a+4 d=45$\ldots(i)
and $11^{\text {th }}$ term of the corresponding
A.P. $=a+10 d=69$\ldots(ii)
From (i) and (ii), we get$a=29, d=4$
Therefore $16^{\text {th }}$ term of the corresponding A.P.
$=a+15 d=29+15 \times 4=89$
Hence $16^{\text {th }}$ term of the H.P. is $\frac{1}{89}$
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