Question

the sum of the 2nd and the 7th term of an AP is 30. if its 15th term is 1 less than twice its 8th term.Find the AP

Answer 1

$a2 + a7 = 30 \\ a + d + a + 6d = 30 \\ 2a + 7d = 30.........1st \: equation \\ a15 = 2a8 - 1 \\ a + 14 = 2a - a - 1 \\ 7d = a - 1 \\ a - 7d = 1 \\ ...........2nd \: equation \\ by \: elimination \: method \: of \: two \: equations \\ we \: get \\ d = 4 \div 3 \\ a - 7d = 1 \\ a - 7(4 \div 3) = 1 \\ a - 28 \div 3 = 1 \\ a = 1 + 28 \div 3 \\ a1 = 31 \div 3 \\ a2 = 31 \div 3 + 4 \div 3 = 35 \div 3 \\ a3 = a2 + d = 35 \div 3 + 4 \div 3 = 39 \div 3 \\ ap \: is \: 31 \div 3 \: \: 35 \div 3 \: \: 39 \div 3........$
hope this helps you quickly

Answer 2

Here is your solution

Given:-

t2+t7=30
=>a+d+a+6d=30
=>2a+7d=30..................... (1)

A/q
15th term=2×8th term-1
=>a+14d=2×(a+7d)-1
=>a+14d=2a+14d-1
=>14d-14d+1=2a-a
=>1=a

Putting the value of a in equation (1)

2a×7d=30
2×1+7d=30
2+7d=30
7d=30-2
7d=28
d=4

A.p are a,a+d,a+2d,a+3d,a+4d

=>A.p are 1,5,9,13.........

HOPE IT HELPS YOU