heya!!
Ques . If the first , second and last terms of AP are a , b and 2a respectively . Then its sum is
A) ab/2(b-a)
B) ab/( b-a)
C) 3ab / 2 (b-a)
D) None of these
Complete solution ✔✔✔✔✔
not just options otherwise reported ❎❎❎❎❎❎❎❎❎❎❎❎
Thanks
Answers
Answer:
C) 3ab / 2(b-a)
Step-by-step explanation:
a, b, 2a are in AP
=> b - a = 2a - b [ the common difference ]
=> 3a = 2b
The sum of the three numbers is then
S = a + b + 2a = 3a + b = 2b + b = 3b
Writing A for the expression in (A), we have
A = ab / 2(b-a)
= 3ab / 2(3b-3a) [ multiply numerator and denominator by 3 ]
= 2b² / 2(3b-2b) [ use 3a = 2b ]
= 2b² / 2b
= b
and this is NOT equal to the sum S = 3b.
B = ab / (b-a) = 2A = 2b, is also NOT equal to the sum S = 3b.
C = 3A = 3b = S.
So the answer is (C).
Here,
a1 = a
a2 = b
common difference d = a2- a1 =
= b-a
let n be the number of terms in series
an = 2a = a + (n-1)d
(n-1) = a /(b-a)
n = (a/(b-a)) + 1
n= b/(b-a)
: sum = n/2 ( a1 +an)
= [b/2(b-a)] (a + 2a)
= 3ab/2(b-a)
So option c is the answer
Thanks!!!