Question

the sum the sum of six terms which form an ap is 345 the difference between first and last term is 55 find the terms ​

Answer 1

Answer:

30,41,52,63,74 and 85

Step-by-step explanation:

number of terms(n) = 6

Let first term be a.

difference between first and last term = 55

Therefore, last term will be [a + (6 - 1)11] = a + 55

Sum of all terms in A.P. = $\frac{n}{2}[2a + (n - 1)d ]$  where, d = Difference between each consecutive term in A.P.

⇒ $\frac{6}{2} [ 2a + (5)11 ] = \frac{6}{2} [ 2a + 55 ]$  = 345

⇒ 345 / 3 = 2a + 55

⇒ 115 - 55 = 2a

⇒ a = 30

Therefore the required terms are 30, 41, 52, 63, 74 and 85

Hope it helps!!