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
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?
Step-by-step explanation:
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:
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