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How many terms are there in 20, 25, 30 . . . . . . 140?
Answer & Solution
Correct Answer:
Option
B
$$\eqalign{
& {\text{Number}}\,{\text{of}}\,{\text{terms}}, \cr
& = {\frac{{ {{1^{st}}\,t{\text{erm - last}}\,{\text{term}}} }}{{{\text{common}}\,{\text{difference}}}}} + 1 \cr
& = { \frac{{140 -20}}{5}} + 1 \cr
& = {\frac{{120}}{5}} + 1 \cr
& = 24 + 1 \cr
& = 25 \cr} $$
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LoginSo,an=a+(n-1)d
140=20+(n-1)5
120=5n-5
n=125/5
n=25
L = a + (n-1)d where L = last term = 140
a = 20 , d= 5 , n = ?
n -1 = (L - a)/d
n = 25