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

Explanation:

Hola frnd!!!!!!!!

$a = 10 \\ tn = 361 \\ d = 9$

$tn = a + (n - 1)d \\ 361 = 10 + 9n - 9 \\ 361 = 1 + 9n \\ 9n = 360 \\ n = 40 \\ so \: there \: are \: fourty \: terms \: ...$

$sn = \frac{n}{2} (a + tn) \\ \: \: \: \: \: \: = \frac{40}{2} (371)$

$\: \: \: \: = 20 \times 371 \\ \: \: \: \: = 7420 \\ hope \: it \: helps....$