Question

the first and last term of anAPare 17 and 350 respectively if the common difference is 9 how many terms are there and what is their sum​

Answer 1

$\bf\huge\underline{Question}$

The first and last term of an AP are 17 and 350 respectively if the common difference is 9, how many terms are there and what is their sum?

$\bf\huge\underline{Answer}$

We have, first term(a) = 17, last term,(l) = 350 = ${T_n}$ and common difference (d) = 9

Let the number of terms be n.

${T_n}$ = a + (n - 1)d

350 = 17 + (n - 1) × 9

=> ( n - 1) × 9 = 350 - 17 = 333

=> n - 1 = $\dfrac{333}{9}$ = 37

=> n = 37 + 1 = 38

Since, ${S_n}$ = $\dfrac{n}{2}$ (a + l)

${S_3_8}$ = $\dfrac{38}{2}$ (17 + 350) = 19(367) = 6973

Thus, n = 38 and ${S_3_8}$ = 6973

Answer 2

Given :

• First term, a = 17
• Last term, l = 350
• Common difference, d = 9

To Find :

• Number of terms in AP, n = ?
• Sum of total number of terms in AP, $\sf S_{n} = ?$

Solution :

Let, l be the nth term of AP.

$\sf : \implies a_{n} = l = 350$

Now, we know that :

$\Large \underline{\boxed{\bf{ a_{n} = a + ( n - 1 ) d }}}$

By, putting values,

$\sf : \implies 350 = 17 + ( n - 1 ) \times 9$

$\sf : \implies 350 = 17 + 9n - 9$

$\sf : \implies 350 = 8 + 9n$

$\sf : \implies 350 - 8 = 9n$

$\sf : \implies 342 = 9n$

$\sf : \implies \dfrac{ \cancel{342}^{38}}{\cancel{9}} = n$

$\sf : \implies 38 = n$

$\sf : \implies n = 38$

$\large \underline{\boxed{\sf n = 38}}$

Hence, There are 38 number of terms in given AP.

Now, let's find sum of total number of terms in AP.

We know that :

$\Large \underline{\boxed{\bf{ S_{n} = \dfrac{n}{2} ( a + a_{n} ) }}}$

We have :

• n = 38
• a = 17
• $\sf a_{n} = 350$

$\sf : \implies S_{38} = \dfrac{\cancel{38}^{1}}{\cancel{2}} ( 17 + 350 )$

$\sf : \implies S_{38} = 19 (367)$

$\sf : \implies S_{38} = 19 \times 367$

$\sf : \implies S_{38} = 6973$

$\large \underline{\boxed{\sf S_{38} = 6973}}$

Hence, There are 38 number of terms in given AP and their sum is 6973.