Question

If in an AP a=2, d=4, then sum of first 40 terms is

Answer 1

ANSWER:

• sum of 40 terms = 3200

GIVEN:

• a (first term) = 2
• d ( common difference) = 4
• n (number of terms) = 40

TO FIND:

• Sum of first 40 terms.

SOLUTION:

Formula:

Sum of n terms = n(2a+(n-1)d)/2

Here:

a = 2

d= 4

n = 40

Putting the values in the formula we get;

=> sum of 40 terms = 40(2*2+(40-1)4)/2

=> sum of 40 terms = 40(4+39*4)/2

=> sum of 40 terms = 20(4+156)

=> sum of 40 terms = 20(160)

=> sum of 40 terms = 3200

• sum of 40 terms = 3200

NOTE:

Some important formulas:

=> nth term = a+(n-1)d

=>Sum of n terms = n(2a+(n-1)d)/2

Where:

a= first term

d= common difference

n= number of terms

Answer 2

GIVEN:

The first term of an AP = a = 2

The common difference = d = 4

TO FIND:

The sum of first 40 terms of AP

SOLUTION:

We know that,

Sum of n term in an AP = n/2 [ 2a + (n - 1)d ]

Where,

n = number of terms

a = first term of AP

d = Common Difference

Sum of first 40 terms = 40/2 [ 2(2) + (40 - 1)4]

= 20 [ 4 + 39 × 4 ]

= 20 [ 4 + 156 ]

= 20 [ 160 ]

= 3200

Therefore,sum of first 40 terms is 3200.