Question

In an A.P a=2, n=16, s16=392, d=?.​

Answer 1

Answer:

Sn = a+(n-1)d

392= 2+(16-1)d

390 = 15d

d. = 390/15

d = 26

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Answer 2

Step-by-step explanation:

Given :-

In an A.P ,a = 2 ,n=16, s16=392

To find :-

Find the value of d ?

Solution :-

Given that

In an AP ,

First term (a) = 2

number of terms (n) = 16

Let the Common difference be d

Sum of first 16 terms (S16) = 392

We know that

Sum of first n terms in an AP is

Sn = (n/2)[2a+(n-1)d]

On substituting these values in the above formula then

=> S16= (16/2)[2(2)+(16-1)d]

=> (8)[4+(15d)] = 392

=> (8)(4+15d) = 392

=> 4+15d = 392/8

=> 4+15d = 49

=> 15d = 49-4

=> 15d = 45

=> d = 45/15

=> d = 3

Therefore, d = 3

Comm difference of the AP = 3

Answer :-

The value of d for the given problem is 3

Check :-

a = 2

d = 3

n = 16

S16 = (16/2)[2×2+(16-1)(3)]

=> S16 = (8)[4+(15×3)]

=> S16 = (8)(4+45)

=> S16 = (8)(49)

=> S16 = 392

Verified the given relations in the given problem.

Used formulae:-

→ Sum of first n terms in an AP is

Sn = (n/2)[2a+(n-1)d]

• a = First term
• d = Common difference
• n = Number of terms