Question

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:

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Step-by-step explanation:

Given, first term, a = 10

Last term, a1 = 361

And, common difference, d = 9

Now a1 =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

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Answer 2

Answer:

Step-by-step explanation

a of n =a+(n-1)d

361=10+(n-1)9

361-10=9n-9

351+9=9n

360=9n

360/9=n

40=n

No of terms=40

Sum =n/2[2a+(n-1)d]

Sum=40/2[20+(40-1)9]

Sum=20[20+351]

Sum=20 [371]

Sum=7420

Sum of 4o terms is7472