Question

The first and the last terms of an A Pare 17 and 350 respectively. If the common difference
is 9, how many terms are there and what is their sum?​

Answer 1

Answer:

38

Step-by-step explanation:

a=17, Tn=350, d=9

Tn= a+(n-1)d

350=17+(n-1)*9

350-17/9=n-1

333/9=n-1

37=n-1

n=37+1

n=38 Answer

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.