Question

for an ap 10,15,20...195 then find the sum of all its term ​

Answer 1

Answer:

The AP is

10,15, 20.......195

$l = a + (n - 1)d$

L : last term

a : 1st term

d : common difference

n : number of terms

195 = 10+(n-1)5

38 = n

$s = n \div 2(a + l)$

S = 38/2(10+195)

S = 19*205

S = 3895

Hope it helps

Answer 2

Answer:

The sum of all the numbers = 3895

Step-by-step explanation:

10,15,20...195 this series is an arithmetic progression.

a = First term = 10

d = Common difference = 5

n = number of terms

Here 195 is the nth term, then 195 = a + (n − 1) × d

                                                  195 = 10 + 5(n - 1)

                                                  n - 1 = 185 ÷ 5 = 37

                                                  n = 37 + 1 = 38

The sum of these 38 terms = n/2 (first term + last term)

                                             = $\frac{38(10 + 195)}{2}$ = 3895

∴ The sum of all the numbers = 3895.

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