what would be the last term of the arithmetic progression with 10 terms whose second term is -23 and the third term is -35
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Answer:
a=first term
a+d =2nd term
a+2d=3rd term
a+d= -23—eq(i)
a+2d= -35—eq(ii)
solve it by elimination method
by subtracting
a+d=-23
a+2d=-35
- -. +
________
-d=12
d=-12
put value of d in eq(I)
a+(-12)=-23
a-12=-23
a=-23+12
-11
a10=a+(n-1)×d
{where n is number of term}
a10= -11+(10-1)×(-12)
-11+9(-12)
-11+(-108)
-11-108
-119
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