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Question: Find the sum to n terms of the A.P. whose \({{\text{k}}^{th}}\) term is \(5{\text{k + 1}}\)....

Find the sum to n terms of the A.P. whose kth{{\text{k}}^{th}} term is 5k + 15{\text{k + 1}}.

Explanation

Solution

We will form the series of A.P. using the given term. We will put the values of k as 1,2,3,...1,2,3,... to get the terms of the series. Then we will use the formula of sum of n terms of A.P. series which is given as-
Sn=n2[2a + (n1)d]\Rightarrow {{\text{S}}_{\text{n}}} = \dfrac{{\text{n}}}{2}\left[ {2{\text{a + }}\left( {{\text{n}} - {\text{1}}} \right){\text{d}}} \right]
Here n is the number of the terms, ‘a’ is the first term and d is the common difference between the terms of the series. Solve the formed equation and we’ll get the answer.

Complete step-by-step answer:
Given, the kth{{\text{k}}^{th}} term of an arithmetic series is 5k + 15{\text{k + 1}}. We have to find the sum of n terms of this series. So first we will find the series. We can write-
ak=5k + 1\Rightarrow {{\text{a}}_{\text{k}}} = 5{\text{k + 1}} - (i)
On putting k=1 = 1 we get-
a1=5 + 1\Rightarrow {{\text{a}}_1} = 5{\text{ + 1}}
On adding the given terms, we get-
a1=6\Rightarrow {{\text{a}}_1} = 6
On putting k=2 = 2 we get-
a2=(5×2) + 1\Rightarrow {{\text{a}}_2} = \left( {5 \times 2} \right){\text{ + 1}}
On solving, we get-
a2=10+1\Rightarrow {{\text{a}}_2} = 10 + 1
On adding the given terms, we get-
a2=11\Rightarrow {{\text{a}}_2} = 11
On putting k=3 = 3 we get-
a3=(5×3) + 1\Rightarrow {{\text{a}}_3} = \left( {5 \times 3} \right){\text{ + 1}}
On solving, we get-
a3=15+1\Rightarrow {{\text{a}}_3} = 15 + 1
On adding the given terms, we get-
a3=16\Rightarrow {{\text{a}}_3} = 16
Similarly, on continuing, we get a series whose terms are 66,1111,1616,…
In this series, first term a = 6{\text{a = 6}} and common difference d = 11 - 6 = 16 - 11 = 5{\text{d = 11 - 6 = 16 - 11 = 5}}
So this series is in A.P.
Now, we will find the sum of n terms of this series using the formula-
Sn=n2[2a + (n1)d]\Rightarrow {{\text{S}}_{\text{n}}} = \dfrac{{\text{n}}}{2}\left[ {2{\text{a + }}\left( {{\text{n}} - {\text{1}}} \right){\text{d}}} \right]
Here n is the number of the terms, ‘a’ is the first term and d is the common difference between the terms of the series.
On putting the given values in the formula, we get-
Sn=n2[(2×6) + (n1)5]\Rightarrow {{\text{S}}_{\text{n}}} = \dfrac{{\text{n}}}{2}\left[ {\left( {2 \times 6} \right){\text{ + }}\left( {{\text{n}} - {\text{1}}} \right)5} \right]
On solving, we get-
Sn=n2[12 + 5n5]\Rightarrow {{\text{S}}_{\text{n}}} = \dfrac{{\text{n}}}{2}\left[ {12{\text{ + }}5{\text{n}} - 5} \right]
On further solving, we get-
Sn=n2[7 + 5n]\Rightarrow {{\text{S}}_{\text{n}}} = \dfrac{{\text{n}}}{2}\left[ {{\text{7 + }}5{\text{n}}} \right]
The sum of n terms of the given A.P. is n2[7 + 5n]\dfrac{{\text{n}}}{2}\left[ {{\text{7 + }}5{\text{n}}} \right].

Note: We can also solve this question using the formula which gives the relation between the first term and last term of A.P. Here clearly the first term will be 66 and we can find the last term by putting k=n. Then use the formula of sum of n terms of A.P. which is given as-
Sn=n2[a+l]\Rightarrow {{\text{S}}_{\text{n}}} = \dfrac{{\text{n}}}{2}\left[ {a + l} \right]
Put the required values in the above formula to get the answer.