# The third term of an AP is 7 and its ${{\text{7}}^{{\text{th}}}}$ term is 2 more than thrice of its ${{\text{3}}^{{\text{rd}}}}$ term. Find the first term, common difference and the sum of its first 20 terms.

Last updated date: 20th Mar 2023

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Answer

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Hint- Here, we will be using the formulas of an AP series.

Given, third term of AP, \[{a_3} = 7\]

Also, \[{a_7} = 3{a_3} + 2\] where \[{a_3}\] is the third term of AP and \[{a_7}\] is the seventh term of AP

Since, in an AP series, the common difference \[d\] remains the same.

Since, \[{n^{th}}\] term of AP is given by \[{a_n} = {a_1} + \left( {n - 1} \right)d\] where \[{a_1}\] is the first term of AP series and \[d\] is the common difference of AP.

\[

\Rightarrow {a_3} = 7 \Rightarrow {a_1} + \left( {3 - 1} \right)d = 7 \Rightarrow {a_1} + 2d = 7{\text{ }} \to {\text{(1)}} \\

\Rightarrow {a_7} = 3{a_3} + 2 \Rightarrow {a_1} + \left( {7 - 1} \right)d = 3\left[ {{a_1} + \left( {3 - 1} \right)d} \right] + 2 \Rightarrow {a_1} + 6d = 3\left[ {{a_1} + 2d} \right] + 2 \\

\Rightarrow {a_1} + 6d = 3{a_1} + 6d + 2 \Rightarrow {a_1} = 3{a_1} + 2 \Rightarrow 2{a_1} = - 2 \Rightarrow {a_1} = - 1 \\

\]

Put this value of \[{a_1}\] in equation (1), we get

\[ - 1 + 2d = 7 \Rightarrow 2d = 8 \Rightarrow d = \frac{8}{2} = 4\]

As we know that sum of first \[n\] terms in AP is given by \[{{\text{S}}_n} = \frac{n}{2}\left[ {2{a_1} + \left( {n - 1} \right)d} \right]\]

Let’s substitute the values of \[{a_1}\] and \[d\], we get

Now, sum of its first 20 terms \[{{\text{S}}_{20}} = \frac{{20}}{2}\left[ {2 \times \left( { - 1} \right) + 4\left( {20 - 1} \right)} \right] = 10\left( { - 2 + 76} \right) = 740\].

Therefore, the first term of the given AP series with common difference 4 is \[ - 1\] and the sum of its first 20 terms is 740.

Note- In these types of problems, find the common parameters which includes the first term, common difference and the total number of terms in the AP series using the given data and then find whatever is asked in the problem.

Given, third term of AP, \[{a_3} = 7\]

Also, \[{a_7} = 3{a_3} + 2\] where \[{a_3}\] is the third term of AP and \[{a_7}\] is the seventh term of AP

Since, in an AP series, the common difference \[d\] remains the same.

Since, \[{n^{th}}\] term of AP is given by \[{a_n} = {a_1} + \left( {n - 1} \right)d\] where \[{a_1}\] is the first term of AP series and \[d\] is the common difference of AP.

\[

\Rightarrow {a_3} = 7 \Rightarrow {a_1} + \left( {3 - 1} \right)d = 7 \Rightarrow {a_1} + 2d = 7{\text{ }} \to {\text{(1)}} \\

\Rightarrow {a_7} = 3{a_3} + 2 \Rightarrow {a_1} + \left( {7 - 1} \right)d = 3\left[ {{a_1} + \left( {3 - 1} \right)d} \right] + 2 \Rightarrow {a_1} + 6d = 3\left[ {{a_1} + 2d} \right] + 2 \\

\Rightarrow {a_1} + 6d = 3{a_1} + 6d + 2 \Rightarrow {a_1} = 3{a_1} + 2 \Rightarrow 2{a_1} = - 2 \Rightarrow {a_1} = - 1 \\

\]

Put this value of \[{a_1}\] in equation (1), we get

\[ - 1 + 2d = 7 \Rightarrow 2d = 8 \Rightarrow d = \frac{8}{2} = 4\]

As we know that sum of first \[n\] terms in AP is given by \[{{\text{S}}_n} = \frac{n}{2}\left[ {2{a_1} + \left( {n - 1} \right)d} \right]\]

Let’s substitute the values of \[{a_1}\] and \[d\], we get

Now, sum of its first 20 terms \[{{\text{S}}_{20}} = \frac{{20}}{2}\left[ {2 \times \left( { - 1} \right) + 4\left( {20 - 1} \right)} \right] = 10\left( { - 2 + 76} \right) = 740\].

Therefore, the first term of the given AP series with common difference 4 is \[ - 1\] and the sum of its first 20 terms is 740.

Note- In these types of problems, find the common parameters which includes the first term, common difference and the total number of terms in the AP series using the given data and then find whatever is asked in the problem.

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