Question

If the $n^{th}$ term of the series 25 + 29 + 33 + 37+ …. And 3 + 4 + 6 + 9 + 13 …. are equal, then what is the value of n?
(a) 11
(b) 12
(c) 13
(d) 14

Answer 1

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.

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,
$\begin{array}{rl}& a_n=25+(n-1)\times 4\\ & a_n=4n+21....(1)\end{array}$
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 $a,bandc$ are any three random constants now, and we are given that
$\begin{array}{rl}& a_1=a+b+c=3\\ & a_2=4a+2b+c=4\\ & a_3=9a+3b+c=6\end{array}$
By solving above three equations, we get
$a={\displaystyle \frac{1}{2}},b=-{\displaystyle \frac{1}{2}}andc=3.$
Hence, We get $n^{th}$ term of the second series as,
$a_n={\displaystyle \frac{n^2-n+6}{2}}.....(2)$
Now by equating equations (1) and (2), we get
$4n+21={\displaystyle \frac{n^2-n+6}{2}}$
After cross-multiplying, we have
$\begin{array}{rl}& n^2-n+6=8n+42\\ & n^2-9n-36=0\\ & (n-12)(n+3)=0\end{array}$
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.

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 $a,bandc$are any three random constants.