What is the 15th term of the series \[3,{\rm{ }}7,{\rm{ }}13,{\rm{ }}21,{\rm{ }}31,{\rm{ }}43\], .....?
A) \[205\]
B) \[225\]
C) \[238\]
D) \[241\]

Hint: In this question we have to find 15th term of given series. Here we have to first find the pattern of the series. Find difference of each consecutive term in the series then observe the pattern. Then follow the same pattern and use appropriate formula to get the required value.

Formula used: \[{S_n} = \dfrac{n}{2}(2a + (n - 1)d)\]
Where
\[{S_n}\]Sum of n terms of AP
n is number of terms
a is first term
d is common difference

Complete step by step solution: Given: series \[3,{\rm{ }}7,{\rm{ }}13,{\rm{ }}21,{\rm{ }}31,{\rm{ }}43\]…
Now add all these numbers
\[s = 3 + 7 + 13 + 21 + 31 + ...... + {a_n}\]……………………………….. (i)
\[ - s = - 3 - 7 - 13 - 21 - ......{a_{n - 1}} - {a_n}\]………………………… (ii)
Add equation (i) and (ii)
\[{a_n} = 3 + 4 + 6 + 8 + 10 + 12......(n - 1)\]Term
Now \[4 + 6 + 8 + 10 + 12\]… all these terms are in AP and \[2\] is a common difference
Sum of all terms is
\[{S_n} = \dfrac{n}{2}(2a + (n - 1)d)\]
\[{a_n} = 3 + \dfrac{{(n - 1)}}{2}(8 + 2n - 4)\]
\[{a_n} = 3 + \dfrac{{(n - 1)}}{2}(4 + 2n)\]
\[{a_n} = 3 + (n - 1)(2 + n)\]
In order to find 15th terms put n equal to 15
\[{a_{15}} = 3 + (15 - 1)(2 + 15)\]
\[{a_{15}} = 3 + 14 \times 17\]
\[{a_{15}} = 3 + 238\]
Now required value is
\[{a_{15}} = 241\]

Thus, Option (D) is correct.

Note: Whenever given series doesn’t follow any pattern then we have to check pattern in difference of each consecutive term. If we get any pattern then follow that pattern to get required values. Don’t try to find first term and common ratio in this type of questions try to find the relation between nth term and given series.
Sometime students get confused in between AP and GP the only difference in between them is in AP we talk about common difference whereas in GP we talk about common ratio.