Question

the first and last term of an AP are 6 and 348 respectively if its common differnce is 9 how many terms are there and what is there sum

Answer 1

348=6+(n-1)6
342=(n-1)6
n=342/6  +1


n=39//

Answer 2

The total number of terms are 39 and the sum of the terms are 6903 when an AP's common difference is 9 and its first and last terms are 6 and 348.

Given that,

We have to find if an AP's common difference is 9 and its first and last terms are 6 and 348, respectively, how many terms are there and what is their sum.

We know that,

nth term of arithmetic progression is aₙ = a+(n-1)d

Where,

aₙ is last term of AP

a is first term of AP

d is common difference

n is total term in AP.

So,

a = 6, aₙ = 348, d=9

We get,

348 = 6+(n-1)9

348 = 6+9n-9

348 = 9n-3

348+3 = 9n

351 = 9n

n= $\frac{351}{9}$

n= 39

Now,

Sₙ = $\frac{n}{2}$[2a+(n-1)d]

S₃₉ = $\frac{39}{2}$[2(6)+(39-1)9]

S₃₉ =  $\frac{39}{2}$[12+38×9]

S₃₉ = $\frac{39}{2}$[12+342]

S₃₉ =  $\frac{39}{2}$[354]

S₃₉ = 39×177

S₃₉ = 6903

Therefore, The total number of terms are 39 and the sum of the terms are 6903 when an AP's common difference is 9 and its first and last terms are 6 and 348.

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