
The sum of the first 9 terms of an A.P is 81 and the sum of its first 20 terms is 400. Find the first term, the common difference and the sum up to 15th term.
Answer
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Hint: The sum of 9 terms which are in A.P is given to us and the sum of the first 200 terms is also given. We need to find the first term, common difference and the sum up to the first 15th term. Directly use the basic formula for n terms of an A.P and form two different equations using the given conditions.
Complete step-by-step answer:
As we know the formula of sum of an A.P is ${S_n} = \dfrac{n}{2}\left( {2a + \left( {n - 1} \right)d} \right)$ where n is the number of terms, a is the first term and d is the common difference.
And it is given that the sum of the first 9 terms of an A.P is 81.
$ \Rightarrow {S_9} = 81 = \dfrac{9}{2}\left( {2a + \left( {9 - 1} \right)d} \right)$
Now simplify the above equation we have,
$ \Rightarrow 2a + 8d = 18$…………………… (1)
Now it is also given that the sum of the first 20 terms of an A.P is 400.
$ \Rightarrow {S_{20}} = 400 = \dfrac{{20}}{2}\left( {2a + \left( {20 - 1} \right)d} \right)$
Now simplify the above equation we have,
$ \Rightarrow 2a + 19d = 40$……………………… (2)
Now subtract equation (2) from equation (1) we have,
$ \Rightarrow 2a + 19d - 2a - 8d = 40 - 18$
$ \Rightarrow 11d = 22$
$ \Rightarrow d = 2$
Now substitute this value in equation (1) we have,
$ \Rightarrow 2a + 8\left( 2 \right) = 18$
$ \Rightarrow 2a = 18 - 16 = 2$
$ \Rightarrow a = 1$.
Now we have to calculate the sum up to 15th term
${S_{15}} = \dfrac{{15}}{2}\left( {2\left( 1 \right) + \left( {15 - 1} \right)2} \right)$
Now simplify the above equation we have,
${S_{15}} = \dfrac{{15}}{2}\left( {2 + 28} \right) = \dfrac{{15}}{2}\left( {30} \right) = 15\left( {15} \right) = 225$
So the first term of an A.P is 1, common difference is 2, and the sum up to 15th term is 225.
So, this is the required answer.
Hence option (D) none of these is correct.
Note: Whenever we face such types of problems the key concept is simply to have the gist of the basic formula for terms involving A.P. A series is said to be in A.P or arithmetic progression if the common difference remains constant throughout the series. Use these concepts, it will help you get on the right track to reach the solution.
Complete step-by-step answer:
As we know the formula of sum of an A.P is ${S_n} = \dfrac{n}{2}\left( {2a + \left( {n - 1} \right)d} \right)$ where n is the number of terms, a is the first term and d is the common difference.
And it is given that the sum of the first 9 terms of an A.P is 81.
$ \Rightarrow {S_9} = 81 = \dfrac{9}{2}\left( {2a + \left( {9 - 1} \right)d} \right)$
Now simplify the above equation we have,
$ \Rightarrow 2a + 8d = 18$…………………… (1)
Now it is also given that the sum of the first 20 terms of an A.P is 400.
$ \Rightarrow {S_{20}} = 400 = \dfrac{{20}}{2}\left( {2a + \left( {20 - 1} \right)d} \right)$
Now simplify the above equation we have,
$ \Rightarrow 2a + 19d = 40$……………………… (2)
Now subtract equation (2) from equation (1) we have,
$ \Rightarrow 2a + 19d - 2a - 8d = 40 - 18$
$ \Rightarrow 11d = 22$
$ \Rightarrow d = 2$
Now substitute this value in equation (1) we have,
$ \Rightarrow 2a + 8\left( 2 \right) = 18$
$ \Rightarrow 2a = 18 - 16 = 2$
$ \Rightarrow a = 1$.
Now we have to calculate the sum up to 15th term
${S_{15}} = \dfrac{{15}}{2}\left( {2\left( 1 \right) + \left( {15 - 1} \right)2} \right)$
Now simplify the above equation we have,
${S_{15}} = \dfrac{{15}}{2}\left( {2 + 28} \right) = \dfrac{{15}}{2}\left( {30} \right) = 15\left( {15} \right) = 225$
So the first term of an A.P is 1, common difference is 2, and the sum up to 15th term is 225.
So, this is the required answer.
Hence option (D) none of these is correct.
Note: Whenever we face such types of problems the key concept is simply to have the gist of the basic formula for terms involving A.P. A series is said to be in A.P or arithmetic progression if the common difference remains constant throughout the series. Use these concepts, it will help you get on the right track to reach the solution.
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