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The first term and common difference of an A.P. is 10 and 5 respectively. Find the sum of the first 30 terms of the A.P.

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Last updated date: 13th Jul 2024
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Answer
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Hint: Use the formula of the sum of an A.P. and put the given data in the formula to reach the final answer.

Complete step-by-step answer:

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 ${{S}_{r}}$.
${{T}_{r}}=a+\left( r-1 \right)d$
${{S}_{r}}=\dfrac{r}{2}\left( 2a+\left( r-1 \right)d \right)$
Now moving to find the sum of the arithmetic progression. The sum of a sequence is generally denoted by ${{S}_{n}}$ . It is given in the question that the first term of the arithmetic progression is 10 and the common difference is 5.
We know that sum of k terms of an arithmetic progression is equal to $\dfrac{k}{2}\left( 2a+(k-1)d \right)$, where a is the first term of the arithmetic progression and d represents the common difference.
\[\therefore {{S}_{30}}=\dfrac{30}{2}\left( 2\times 10+\left( 30-1 \right)5 \right)\]
\[\Rightarrow {{S}_{30}}=15\left( 20+29\times 5 \right)=165\times 15=2475\]
Therefore, the sum of the first 30 terms of an arithmetic progression whose first term is 10, and the common difference is 5 is 2475.

Note: Be careful about the signs and try to keep the equations as neat as possible by removing the removable terms. Moreover, we should know the formulas related to arithmetic progressions and geometric progressions as they are used quite often.