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?
Answers
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
Step-by-step explanation:
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