How many terms are there in 20, 25, 30 . . . . . . 140?
A. 22
B. 25
C. 23
D. 24
Answer: Option B
Solution (By Examveda Team)
$$\eqalign{ & {\text{Number}}\,{\text{of}}\,{\text{terms}}, \cr & = {\frac{{ {{1^{st}}\,t{\text{erm - last}}\,{\text{term}}} }}{{{\text{common}}\,{\text{difference}}}}} + 1 \cr & = { \frac{{140 -20}}{5}} + 1 \cr & = {\frac{{120}}{5}} + 1 \cr & = 24 + 1 \cr & = 25 \cr} $$This Question Belongs to Arithmetic Ability >> Progressions
can we calculate without any formula
GOOD
Zs,,s
Very nice
Here a=20,an=140,d=5
So,an=a+(n-1)d
140=20+(n-1)5
120=5n-5
n=125/5
n=25
Some mistake in calculation
L = a + (n-1)d where L = last term = 140
a = 20 , d= 5 , n = ?
n -1 = (L - a)/d
n = 25
25....