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:

• First term ( a ) = 10
• Last term ( l ) = 361
• Common difference ( d )= 9

• Calculation of Number of Terms :

⇒ l = a + (n −1)d

⇒ 361 = 10 + (n − 1)9

⇒ 361 = 10 + 9n − 9

⇒ 361 = 9n + 1

⇒ 9n = 360

⇒ n = 40

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 × 9]

⠀⠀= 20 [20 + 351]

⠀⠀ = 20 × 371 = 7420

∴ Sum of all 40 terms of AP is 7420.