Question

Let \[{{a}_{1}},{{a}_{2}},.........,{{a}_{30}}\] be an A.P. \[S=\sum\limits_{i=1}^{30}{{{a}_{i}}}\] and \[T=\sum\limits_{i=1}^{15}{{{a}_{\left( 2i-1 \right)}}}\]. If \[{{a}_{5}}=27\] and \[S-2T=75\], then \[{{a}_{10}}\] is equal to ?

Answer 1

Hint: In the given question, we have been asked the tenth term of an AP. In order to find, we first need to find the common difference of an AP. Then we find the first term of an AP and after that put the value of ‘a’ and ‘d’, we get \[{{10}^{th}}term\ i.e.\ {{a}_{10}}\] of an AP.

Complete step by step solution:
Any \[{{n}^{th}}\]term of an AP i.e. arithmetic progression is,
 \[{{a}_{n}}=a+\left( n-1 \right)d\], where
\[a=\] First term of an AP
\[n=\ {{n}^{th}}\]term of an AP
d = common difference between two consecutive term of an AP
\[sum\ of\ {{n}^{th}}\ term\ of\ an\ AP=\dfrac{n}{2}\left( first\ term+last\ term \right)\]
Let the common difference be ‘d’, then
\[\Rightarrow {{a}_{1}}+{{a}_{2}}+{{a}_{3}}+...........+{{a}_{30}}=S=\dfrac{30}{2}\left( {{a}_{1}}+{{a}_{1}}+29\times d \right)=\dfrac{30}{2}\left( 2{{a}_{1}}+29d \right)\]
Also,
\[\Rightarrow {{a}_{1}}+{{a}_{3}}+{{a}_{5}}+{{a}_{7}}...........+{{a}_{29}}=T=\dfrac{15}{2}\left( 2{{a}_{1}}+28\times d \right)\]
Now, it has given that
\[\Rightarrow {{a}_{5}}=27\]
\[\Rightarrow a+4d=27\]----- (1)
Also, it has given that
\[\Rightarrow S-2T=75\]
Putting the value of \[S=\dfrac{30}{2}\left( 2{{a}_{1}}+29\times d \right)\] and \[T=\dfrac{15}{2}\left( 2{{a}_{1}}+28\times d \right)\] in the above equation, we get
\[\Rightarrow ~~\dfrac{30}{2}\left( 2{{a}_{1}}+29d \right)-2\times \dfrac{15}{2}\left( 2{{a}_{1}}+28d \right)=75\]
\[\Rightarrow ~~15\left( 2{{a}_{1}}+29d \right)-15\left( 2{{a}_{1}}+28d \right)=75\]
Simplifying the above, we get
\[\Rightarrow 15\times d=75\]
Divide both the sides of the above equation by 15, we get
\[\Rightarrow \dfrac{15\times d}{15}=\dfrac{75}{15}\]
Solving the above equation, we get
\[\therefore d=5\]
Putting the value of ‘d’ = 5 in equation (1), we obtain
\[\Rightarrow a+4\times 5=27\]
Simplifying the above equation, we get
\[\Rightarrow a+20=27\]
Subtract 20 from both the sides of the equation, we get
\[\Rightarrow a+20-20=27-20\]
\[\therefore a=7\]
Thus,
\[\Rightarrow {{a}_{10}}=a+\left( 10-1 \right)\times d=7+\left( 9\times 5 \right)=7+45=52\]
\[\therefore {{a}_{10}}=52\]

Formula used:
Any \[{{n}^{th}}\]term of an AP i.e. arithmetic progression is,
 \[{{a}_{n}}=a+\left( n-1 \right)d\], where
\[a=\] First term of an AP
\[n=\ {{n}^{th}}\]term of an AP
d = common difference between two consecutive term of an AP
\[sum\ of\ {{n}^{th}}\ term\ of\ an\ AP=\dfrac{n}{2}\left( first\ term+last\ term \right)\]

Note: When the numbers have been arranged in a particular manner or order, then the numbers are said to be in sequence. AP i.e. arithmetic progression is a sequence when we add a fixed number to a number to get the next consecutive number. For example if the fixed number is ‘d’ and ‘a’ is the term in the sequence, to get the number next to ‘a’, we will simply add ‘d’ to ‘a’ i.e. a+d is the next term to ‘a’.