Question

In an A.P n=10, d=5, s10=255, a=?​

Answer 1

Step-by-step explanation:

Given :-

In an A.P n=10, d=5, s10=255,

To find :-

Find the value of a ?

Solution :-

Given that

In an AP ,

number of terms (n) = 10

Common difference (d) = 5

Sum of first 10 terms (S10) = 255

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

=> S10 = (10/2)[2a+(10-1)(5)]

=> (5)[2a+(9)(5)] = 255

=> (5)(2a+45) = 255

=> 2a+45 = 255/5

=> 2a+45 = 51

=> 2a = 51-45

=> 2a = 6

=> a = 6/2

=> a = 3

Therefore, a = 3

First term of the AP =

Answer :-

The value of a for the given problem is 3

Check :-

a = 3

d = 5

n = 10

S10 = (10/2)[2×3+(10-1)(5)]

=> S10 = (5)[6+9(5)]

=> S10 = (5)(6+45)

=> S10 = (5)(51)

=> S10 = 255

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