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Question: If the \({{n}^{th}}\) term of an A.P. is given by \({{t}_{n}}=3-4n\), find the sum of the first sixt...

If the nth{{n}^{th}} term of an A.P. is given by tn=3−4n{{t}_{n}}=3-4n, find the sum of the first sixteen terms of the sequence.

Explanation

Solution

First of all find the first two terms of the A.P by substituting n = 1 and n = 2 in the given expression one by one. Now, subtract the first term with the second term to get the common difference. Assume the first term of the A.P as ‘a’, the common difference as d and the number of terms whose sum is to be found as n, apply the formula Sn=n2[2a+(n−1)d]{{S}_{n}}=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right] to get the answer. Here Sn{{S}_{n}} denotes the sum of n terms of an A.P.

Complete step by step solution:
Here we have been provided with the general expression for the nth{{n}^{th}} term of an A.P. given tn=3−4n{{t}_{n}}=3-4n and we are asked to determine its sum of first 16 terms. First we need to find the common difference of this A.P.
Now, to find the common difference of the A.P we need two of its consecutive terms whose difference can be taken, so let us find the first two terms of the A.P. Substituting n= 1 and n = 2 in the expression tn=3−4n{{t}_{n}}=3-4n one by one we get,
t1=3−4(1)=−1 t2=3−4(2)=−5 \begin{aligned} & {{t}_{1}}=3-4\left( 1 \right)=-1 \\\ & {{t}_{2}}=3-4\left( 2 \right)=-5 \\\ \end{aligned}
Subtracting t1{{t}_{1}} from t2{{t}_{2}} we get,
⇒d=−5−(−1) ⇒d=−5+1 ⇒d=−4 \begin{aligned} & \Rightarrow d=-5-\left( -1 \right) \\\ & \Rightarrow d=-5+1 \\\ & \Rightarrow d=-4 \\\ \end{aligned}
Here (d) denotes the common difference.
We know that the sum of n terms of an A.P is given by the relation Sn=n2[2a+(n−1)d]{{S}_{n}}=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right] where n is the number of terms and Sn{{S}_{n}} denotes the sum, so substituting n = 16 we get in the sum formula we get,
S16=162[2×(−1)+(16−1)(−4)] S16=8[−62] S16=−496 \begin{aligned} & {{S}_{16}}=\dfrac{16}{2}\left[ 2\times \left( -1 \right)+\left( 16-1 \right)\left( -4 \right) \right] \\\ & {{S}_{16}}=8\left[ -62 \right] \\\ & {{S}_{16}}=-496 \\\ \end{aligned}
Hence the required sum is -496.

Note: You can also find the answer without the help of common difference by directly using the formula Sn=n2[t1+tn]{{S}_{n}}=\dfrac{n}{2}\left[ {{t}_{1}}+{{t}_{n}} \right]. What you can do is, find the first term and then sixteenth term and use the formula S16=162[t1+t16]{{S}_{16}}=\dfrac{16}{2}\left[ {{t}_{1}}+{{t}_{16}} \right] to get the answer. Remember both the formulas of the sum of terms of A.P as anyone can be used as per the relative ease of application.