The first and last terms of an AP are 7 and 49 respectively. If sum of all terms is 420, find its common difference.​

a=7 l=49 Sn=420

Sn=

2

n

[a+l]

So 420×2=n[7+49]

n=15

l=a+(n−1)d

⇒ 49=7+14 d

⇒ 7=1+2 d ⇒ 2d=6

⇒ d=3

Step-by-step explanation:

First term of AP is 7.

Last term of AP is 49.

Sum of all terms of AP is 420.

What is the common difference of the AP ?

Solution: Let the common difference and number of terms of AP be d and n respectively.

a = first term = 7

d = common difference

l = last term = 49

Sⁿ = sum of all terms = 420

As we know that , Sum of n term of an AP is

⟹ 420 = n/2 ( 7 + 49 )

⟹ 420 × 2 = 56n

⟹ 840/56 = n

⟹ 15 = n

So the given AP series contains 15 terms and last term of AP will be given by -

➙ a + (n – 1)d = 49

➙ 7 + (15 – 1)d = 49

➙ 14d = 49 – 7

➙ d = 42/14

➙ d = 3

Hence, the common difference of AP series is 3.