Question

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

Answer:

Given, first term, a = 10

Last term, al = 361

And, common difference, d = 9

Now al =a + (n −1)d

⟹ 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

Answer 2

Answer:

Let the number of terms be 'n'

So, the 'n'th term is the last term = 361

'n'th term = a + (n - 1)d

where a is the first term and d is the common difference.

so,

361 = 10 + (n-1) 9

361 = 10 + 9n - 9

361 = 9n + 1

9n = 361 - 1 = 360

n = 360/9 = 40

so, there are 40 terms

sum of all 40 terms = (n/2)(a + a40)

= (40/2)(10 + 361)

= 20(371) = 7420

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