Question

If ${{s}_{5}}=15$ and ${{S}_{6}}=21$ then find ${{t}_{6}}$

Answer 1

Hint: We will use the basic definition of sum. Here the sum of first five terms is given to us as 15
And the sum of the first 6 terms is given to us as 21. Now the sum of the first six terms is nothing but the sum of the first five terms + sixth term. Hence we will solve this to get a sixth term.

Complete step-by-step answer:
Now we know that ${{t}_{n}}$ means the nth term of a sequence and ${{S}_{n}}$ means the sum of first n terms of sequence.
Now let the terms of sequence be ${{t}_{1}},{{t}_{2}},{{t}_{3}}....$
Now it is given that ${{S}_{5}}=15$
Here the first five terms of sequence are ${{t}_{1}},{{t}_{2}},{{t}_{3}},{{t}_{4}},{{t}_{5}}$
Hence we have ${{t}_{1}}+{{t}_{2}}+{{t}_{3}}+{{t}_{4}}+{{t}_{5}}=15..............(1)$
Now it is also given to us that ${{S}_{6}}=21$.
Here the first six terms of sequence are ${{t}_{1}},{{t}_{2}},{{t}_{3}},{{t}_{4}},{{t}_{5}},{{t}_{6}}$
Hence we get ${{t}_{1}}+{{t}_{2}}+{{t}_{3}}+{{t}_{4}}+{{t}_{5}}+{{t}_{6}}=21.............(2)$
Now subtracting equation (1) from equation (2) we get
${{t}_{1}}+{{t}_{2}}+{{t}_{3}}+{{t}_{4}}+{{t}_{5}}+{{t}_{6}}-({{t}_{1}}+{{t}_{2}}+{{t}_{3}}+{{t}_{4}}+{{t}_{5}})=21-15$
Here all the terms on the left hand side will be cancelled out and only ${{t}_{6}}$ will remain.
Hence we get ${{t}_{6}}=6$
Hence the value of ${{t}_{6}}=6$

Note: We can directly use the formula ${{t}_{n}}={{S}_{n}}-{{S}_{n-1}}$ to find ${{t}_{n}}$where ${{t}_{n}}$ is the nth and ${{S}_{n}}$ is the sum of first n terms, similarly ${{S}_{n-1}}$ is sum of first n-1 terms. This formula is true for all series. Hence there is no condition for this formula. Alternatively we can also solve this equation by using the formula for Sum of AP. We know the sum of n terms of AP with a first term a and common difference d is given by ${{S}_{n}}=\dfrac{n}{2}[2a+(n-1)d]$ . Hence with this formula we can say $15=\dfrac{5}{2}[2a+(5-1)d]=\dfrac{5}{2}[2a+(4)d]$ and $21=\dfrac{6}{2}[2a+(6-1)d]=\dfrac{6}{2}[2a+5d]$. Hence we have now two equations in two variables. We can solve them to get a and d. and hence then find the 6th term with the formula ${{t}_{n}}=a+(n-1)d$ .