Question

1,11,21...then a20 ....?plz solve this sum it's std 10th sum chapter number 5 ​

Answer 1

Answer:

Step-by-step explanation:

Answer : Given : an AP has 21 terms. the sum of it's 10th, 11th,12th terms is 129 and the sum of the last three terms is 237

To find : AP

 

Let a and d be the first term and common difference of the given A.P.

a 10 + a 11 + a 12 = 129

=> (a + 9d) + (a + 10d) + (a + 11d) = 129 [an = a + (n – 1)d]

=> 3a + 30d = 129

=> a + 10d = 43 … (1)

a 19 + a 20 + a 21 = 237

=> (a + 18d) + (a + 19d) + (a + 20d) = 237

=> 3a + 57d = 237

=> a + 19d = 79 … (2)

Solving (1) and (2), we get

a + 19d – a – 10d = 79 – 43

=> 9d = 36

=> d = 4

When d = 4, we get

a + 10 * 4 = 43

=> a = 43 – 40 = 3

Thus, the given A.P. is 3, 7, 11, 15… Answer

Answer 2

Answer:

Sorry don't know..........