Question

sum of 6 terms of an A P is 90 sum first two term is diminished by 32 with the the sum of last two terms find the A P​

Answer 1

Answer:

sorry

Step-by-step explanation:

i forgot the formula of AP

sorry

Answer 2

The AP is 5,9,13,17,21,25....

Step-by-step explanation:

let the first term be a

and common difference be d

according to question

sum of the first 6 terms is 90

i.e

$S_n =\frac{n}{2}(2a+(n-1)d)\\\\S_6 = \frac{6}{2}(2a+5d)=90\\\\= 3(2a+5d)=90\\\\2a+5d= 30..(1)$

and sum first two term is diminished by 32 with the the sum of last two terms

i.e

sum of first two terms +32 = sum of last two terms

$a+(a+d)+32=a+(6-1)d+(a+(6-1-1)d)\\\\2a+d+32=2a+9d\\\\d=4$

solving 1 we get

2a +5d = 30

2a = 30-20=10

a = 5

therefore , a = 5, d= 4

hence , The AP is 5,9,13,17,21,25

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