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# Find the sum of 2, 7, 12….… to 10 terms:(a) 160(b) 245(c) 290(d) 300

Last updated date: 17th Jun 2024
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Hint: Notice that the terms of the series given in the question are in arithmetic progression with common difference and the first term equal to 5 and 2, respectively. So, directly use the formula of sum of an A.P. to get the answer. The sum of first r terms of an AP is given by ${{S}_{r}}=\dfrac{r}{2}\left( 2a+\left( r-1 \right)d \right)$ , where a is the first term and d is the common difference of the AP.

Before starting with the solution, let us discuss what an A.P. is. A.P. stands for arithmetic progression and is defined as a sequence of numbers for which the difference of two consecutive terms is constant. The general term of an arithmetic progression is denoted by ${{T}_{r}}$ , and sum till r terms is denoted by ${{S}_{r}}$ .
${{T}_{r}}=a+\left( r-1 \right)d$
${{S}_{r}}=\dfrac{r}{2}\left( 2a+\left( r-1 \right)d \right)=\dfrac{r}{2}\left( a+l \right)$
Now to find the sum of the A.P., we will use the formula ${{S}_{r}}=\dfrac{r}{2}\left( 2a+\left( r-1 \right)d \right)$ . In the formula a is the first term, so a=2 and d is the common difference, d=5. Therefore, the sum of the series 2, 7, 12….… to 10 terms is:
${{S}_{r}}=\dfrac{r}{2}\left( 2a+\left( r-1 \right)d \right)$
$\Rightarrow {{S}_{r}}=\dfrac{10}{2}\left( 2\times 2+5\left( 10-1 \right) \right)$
$\Rightarrow {{S}_{r}}=5\left( 4+5\times 9 \right)$
$\Rightarrow {{S}_{r}}=5\times 49=245$