What is the sum of the arithmetic sequence 3, 9, 15..., if there are 26 terms?

Question

surjithayer · Answer

first term a=3
c.d. d-9-3=6
no. of terms =26
$$s _{n}=\frac{ n }{ 2 }[2a+(n-1)d]$$
$$s _{26}=\frac{ 26 }{ 2 }[2(3)+(26-1)(6)]$$
$$s _{26}=13[6+25\times 6]$$
$$s _{26}=13\times 156$$$$=?$$

toga · Answer

The first term of the arithmetic sequence is 3 and the common difference between the terms is 6. To find the sum of an arithmetic sequence, we can use the formula:

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

where S is the sum of the sequence, n is the number of terms, a is the first term, and d is the common difference.

Plugging in the given values, we get:

S = 26/2 [2(3) + (26-1)(6)]
S = 13 [6 + 150]
S = 13 x 156
S = 2028

Therefore, the sum of the arithmetic sequence 3, 9, 15..., if there are 26 terms is 2028.

Manny300303199 · Answer

The sum of the arithmetic sequence 3, 9, 15..., if there are 26 terms is 2028.