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Hint: To solve this question first, we will need to find the ${{n}^{th}}$ term of the each given series as we can observe first given series are in A.P. and we know that the ${{n}^{th}}$ term of the A.P. is given by ${{a}_{n}}=a+(n-1)d$ where a is the first term of the A.P. and d is the common difference of the A.P. similarly to find the ${{n}^{th}}$ term of the second series we can observe that difference of the second series is in A.P. so we can let it’s ${{n}^{th}}$ term as $a{{n}^{2}}+bn+c$ where a, b, c are any 3 arbitrary constants(we know ${{a}_{1}},{{a}_{2\,}}\,and\,{{a}_{3}}$ terms of the series so make equations and solve for ${{n}^{th}}$ term). So, find the ${{n}^{th}}$ term of both series and equate them to find the value of n.

We can observe that the first series is in A.P.,

And we need to find of n for which ${{n}^{th}}$ term of the both given series are equal.

First series we have,

25 + 29 + 33 + 37 + …

Here first term(a) = 25 and common difference(d) = 29 – 25 = 4,

Hence for the ${{n}^{th}}$ term of the first series we get,

$\begin{align}

& {{a}_{n}}=25+(n-1)\times 4 \\

& {{a}_{n}}=4n+21\,\,\,....(1) \\

\end{align}$

Second series we have,

3 + 4 + 6 + 9 + 13 ….

Here we can observe that the difference of the two consecutive terms is in A.P. i.e. 1, 2, 3, 4, … and we know that in that case we need to let the ${{n}^{th}}$ term as $a{{n}^{2}}+bn+c$ where $a,b\,and\,c$ are any three random constants now, and we are given that

$\begin{align}

& {{a}_{1}}=a+b+c=3 \\

& {{a}_{2}}=4a+2b+c=4 \\

& {{a}_{3}}=9a+3b+c=6 \\

\end{align}$

By solving above three equations, we get

$a=\dfrac{1}{2},b=-\dfrac{1}{2}\,and\,c=3.$

Hence, We get ${{n}^{th}}$ term of the second series as,

${{a}_{n}}=\dfrac{{{n}^{2}}-n+6}{2}\,\,.....(2)$

Now by equating equations (1) and (2), we get

$4n+21=\dfrac{{{n}^{2}}-n+6}{2}$

After cross-multiplying, we have

$\begin{align}

& {{n}^{2}}-n+6=8n+42 \\

& {{n}^{2}}-9n-36=0 \\

& (n-12)(n+3)=0 \\

\end{align}$

**Complete step-by-step solution:**We can observe that the first series is in A.P.,

And we need to find of n for which ${{n}^{th}}$ term of the both given series are equal.

First series we have,

25 + 29 + 33 + 37 + …

Here first term(a) = 25 and common difference(d) = 29 – 25 = 4,

Hence for the ${{n}^{th}}$ term of the first series we get,

$\begin{align}

& {{a}_{n}}=25+(n-1)\times 4 \\

& {{a}_{n}}=4n+21\,\,\,....(1) \\

\end{align}$

Second series we have,

3 + 4 + 6 + 9 + 13 ….

Here we can observe that the difference of the two consecutive terms is in A.P. i.e. 1, 2, 3, 4, … and we know that in that case we need to let the ${{n}^{th}}$ term as $a{{n}^{2}}+bn+c$ where $a,b\,and\,c$ are any three random constants now, and we are given that

$\begin{align}

& {{a}_{1}}=a+b+c=3 \\

& {{a}_{2}}=4a+2b+c=4 \\

& {{a}_{3}}=9a+3b+c=6 \\

\end{align}$

By solving above three equations, we get

$a=\dfrac{1}{2},b=-\dfrac{1}{2}\,and\,c=3.$

Hence, We get ${{n}^{th}}$ term of the second series as,

${{a}_{n}}=\dfrac{{{n}^{2}}-n+6}{2}\,\,.....(2)$

Now by equating equations (1) and (2), we get

$4n+21=\dfrac{{{n}^{2}}-n+6}{2}$

After cross-multiplying, we have

$\begin{align}

& {{n}^{2}}-n+6=8n+42 \\

& {{n}^{2}}-9n-36=0 \\

& (n-12)(n+3)=0 \\

\end{align}$

**Hence we get two values of $n$ as 12 and -3 but as $n$ cannot be negative so $n=12$.**

Hence option (b) is the correct answer.Hence option (b) is the correct answer.

**Note:**You need to observe both of the given series carefully in order to solve this problem. And this question involves a lot of calculations so make sure to do that carefully also. And remember whenever difference of the given series is in A.P. then we can let the ${{n}^{th}}$ term of the series as $a{{n}^{2}}+bn+c$ where $a,b\,and\,c$are any three random constants.Recently Updated Pages

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