Find the arithmetic progression whose 4" term is 9 and the sum of its 6" and 13
terms is 40.
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Answer:
A4= 9
A6+A3=40
A4=9
An=a+(n-1)d=9
a+(4-1)d=9
a+3d=9. 1eq
A6+A3=40
a+(n-1)d+a+(n-1)d=40
a+5d+a+2d=40
2a+7d=40
2a=40-7d
a=40-7d/2
put a value in 1eq
40-7d/2+3d/1=9
take LCM of denominators 2 and 1
40-7d+6d/2=9
multiply denominator with RHS value
40-1d=9×2
40-1d=18
-1d=18-40
-1d=-22
d= -22/-1
- and - cancels
d=22
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