Question

For an A.P a=11 and d= 7 Find the sum of first 40term of the A.p​

Answer 1

★ Given that,

• In an AP, a = 11 ; d = 7

★ To find,

• Sum of first 40 terms if AP.

★ Formula :

$\orange{\bigstar}\boxed{\rm{\red{ S_{n} = \cfrac{n}{2}[2a + (n-1) d]}}}\:\orange{\bigstar}$

• Substitute the values.

$\bf\:\implies S_{40} = \cancel{\cfrac{40}{2}} [2(11)+(40-1)7]$

$\bf\:\implies S_{40} =20[22 +(39)7]$

$\bf\:\implies S_{40} =20[22 +273]$

$\bf\:\implies S_{40} =20[295]$

$\bf\:\implies S_{40} = 5900$

$\underline{\boxed{\rm{\purple{\therefore Sum\:of\:first\:40\:terms = 5900.}}}}\:\orange{\bigstar}$

Answer 2

Answer:

5900

Step-by-step explanation:

The formula for finding sum of n terms is :-

Sn = (n/2){2a + (n-1)d}

Given,

a = 11

d = 7

n = 40

Substituting in the formula we get :-

Sn = (40/2){2(11) + (39)(7)}

= (20){22 + 273}

= (20){295}

= 5900