Find the 12th term from the end of the following arithmetic progressions:
(i)3,5,7,9,...201
(ii)3,8,13,...,253
(iii)1,4,7,10,...,88
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Answer:
a) here,
a=201 & d= -2
a12= a + 11d
a12= 201 + 11(-2)
a12= 201 - 22
a12= 189
b) here,
a=253 & d=-5
a12= a + 11d
a12= 253 + 11(-5)
a12= 253 - 55
a12= 198
c) here,
a=88 & d=-3
a12= a + 11d
a12= 88 + 11(-3)
a12= 88 - 33
a12= 55
hope it helps
(please forgive mistakes if any)
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