Question

the sum of 6 terms which form an AP is 345 and the difference between the forest and last term is 55. Find the terms​

Answer 1

Step-by-step explanation:

Let the first term be a

and common difference be d

6th term = a +(6-1)d = a -5d

A/q

(a+5d) -a = 55

⇒5d = 55

⇒d = 11

also, 

S6 = (6/2)[2a+(6-1)×11] = 345

⇒2a +55 = 115

⇒2a = 60

⇒a = 30

so AP = 30,41,52,63,74,85,96.....so on

Answer 2

Terms used

n = no of terms

a = first term

l = last term

Formulae Used

Sum of n terms

$= \frac{n}{2} (a + l)$

nth term of an AP

$= a + (n - 1)d$

SEE THE ATTACHMENT FOR TE REST PART OF THE QUESTION.

So, at the end we have

a = 30 and d = 11

So the terms are :

a , a+d , a+2d , a+3d , a+4d , a+5d

30 , 41 , 52 , 63 , 74 , 85

There's your answer !

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