Question

If the first term and the last term of a finite A.P. are 5 and 95 respectively and $d = 5$ , find n and ${S_n}$ .

Answer 1

For finding n, use the formula ${T_n} = a + \left( {n - 1} \right)d$ . After finding n, find ${S_n}$ using the formula ${S_n} = \dfrac{n}{2}\left( {a + l} \right)$ .

Complete step-by-step answer:
It is given that the first term of an A.P. $a = 5$ , last term $l = 95$ and difference between terms $d = 5$ .
Now, first we have to find $n$ and for that we have to use the formula ${T_n} = a + \left( {n - 1} \right)d$ .

$$\therefore {T_n} = a + \left( {n - 1} \right)d \\\ \therefore 95 = 5 + \left( {n - 1} \right) \times 5 \\\ \therefore 95 = 5 + 5n - 5 \\\ \therefore 5n = 95 \\\ \therefore n = \dfrac{{95}}{5} \\\ \therefore n = 19 \\\$$

Thus, we get a number of terms in the given A.P. as $n = 19$ .
Now, for the sum of the terms, we will use the formula ${S_n} = \dfrac{n}{2}\left( {a + l} \right)$ .
${S_n} = \dfrac{n}{2}\left( {a + l} \right)$
$$
= \dfrac{{19}}{2}\left( {5 + 95} \right) \\
= \dfrac{{19}}{2}\left( {100} \right) \\
= 19 \times 50 \\
= 950 \\

Thus, we get the sum of the terms of the given A.P. as $${S_n} = 950$$ . **Note:** Alternate method for sum: Here, we can also use the formula $${S_n} = \dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]$$ for the sum of terms of given A.P. $$\therefore {S_n} = \dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]$$ $$ = \dfrac{{19}}{2}\left[ {2 \times 5 + \left( {19 - 1} \right) \times 5} \right] \\\ = \dfrac{{19}}{2}\left( {10 + 18 \times 5} \right) \\\ = \dfrac{{19}}{2}\left( {10 + 90} \right) \\\ = \dfrac{{19}}{2}\left( {100} \right) \\\ = 19 \times 50 \\\ = 950 \\\