Question

❣❣please tell me how to solve it...
mean how to start....
❣❣Q. The first and the last terms of an AP are 10 and 361 respectively. If its common difference is 9 then find the number of terms and their total sum?

Answer 1

Heya....❤❤❤

here is ur answer...

Given, first term, a = 10

Last term, al = 361

And, common difference, d = 9

Now al =a + (n −1)
⟹ 361 = 10 + (n − 1)9
⟹ 361 = 10 + 9n − 9
⟹ 361 = 9n + 1
⟹ 9n = 360
⟹ n = 40

Therefore, total number of terms in AP = 40

Now, sum of total number of terms of an AP is given as:

Sn = n/2 [2a + (n − 1)d]
⟹ S40 = 40/2 [2 × 10 + (40 − 1)9]
= 20|20 + 39 x 9]
=20[20 + 351]
=20 × 371 = 7420

Thus, sum of all 40 terms of AP = 7420

hope it helps...

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Answer 2

Given : a = t¹ = 10

tn = 361

d = 9

To find : n and Total sum

Sol :

tn = a + ( n - 1 )d

361 = 10 + ( n - 1 )9

361 = 10 + 9n - 9

361 = 1 + 9n

361 - 1 = 9n

360 = 9n

360÷9 = n

40 = n

Sn = N÷2 ( t1 + tn )

= 40÷2 ( 10 + 361 )

= 20 ( 371 )

= 7420

Therefore, N = 40 and total sum is 7420

if there si something wrong then kindly rectify it.

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