Question

The first and the last term of ap are 8and 350 respectively if its common difference is 9 how many term qre there and what is sum

Answer 1

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S39=6981

Answer 2

Given :

• First term, a = 8
• 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,

$\sf : \implies 350 = 8 + ( n - 1 ) \times 9$

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

$\sf : \implies 350 = - 1 + 9n$

$\sf : \implies 350 + 1 = 9n$

$\sf : \implies 351 = 9n$

$\sf : \implies \dfrac{ \cancel{351}^{39}}{\cancel{9}} = n$

$\sf : \implies 39 = n$

$\sf : \implies n = 39$

$\large \underline{\boxed{\sf n = 39}}$

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 = 39
• a = 8
• $\sf a_{n} = 350$

$\sf : \implies S_{39} = \dfrac{39}{2} ( 8 + 350 )$

$\sf : \implies S_{39} = \dfrac{39}{2} (358)$

$\sf : \implies S_{39} = \dfrac{39}{\cancel{2}} \times \cancel{358}^{179}$

$\sf : \implies S_{39} = 39 \times 179$

$\sf : \implies S_{39} = 6981$

$\large \underline{\boxed{\sf S_{39} = 6981}}$

Hence, There are 39 number of terms in given AP and their sum is 6981.