Quantitative Aptitude – Sequence and Series – if a(base1), a(base2)

Slot – 1 – Quantitative Aptitude – Sequence and Series – if a(base1), a(base2)

if a(base1), a(base2) – Video

Q. If a(base1), a(base2), … are in A.P., then, 1/√a(base1)+√a(base2) + 1/√a(base2)+√a(base3) + … + 1/√a(base n)+√a(base n + 1) is equal to?

n-1/√a(base1)+√a(base n)

n/√a(base1)+√a(base n + 1)

n-1/√a(base1)+√a(base n – 1)

n/√a(base1)-√a(base n + 1)

Answer: n/√a(base1)+√a(base n + 1)

Solution:
The best method to this problem is to use hit and trial method , for this put n =1 in the question you will get the first term only option a) is equal to that so option a) is correct.
Alternative method )
As given a(base1) , a(base2) ,a(base3) , a(base4) , ………..,a(base n) are in AP.
So a(base2) – a(base1) = a(base3) – a(base2)= a(base4) – a(base3) = …….= a(base n) – a(base n-1) = d (common difference)
Now 1/(√a1+√a2) + 1/(√a2+√a3) + 1/(√a3+√a4) + 1/(√a4+√a5)+⋯.
= 1/(√a1+√a2)×(√a1-√a2)/(√a1-√a2)+1/(√a2+√a3)×(√a2-√a3)/(√a2-√a3)+1/(√a3+√a4)×(√a3-√a4)/(√a3-√a4)+⋯.+⋯.1/(√(a(n-1))+√an)×(√an-√(a(n+1)))/(√an-√(a(n+1)))
= (√a1-√a2)/(a1-a2 )+(√a2-√a3)/(a2-a3 )+(√a1-√a2)/(a1-a2 )+⋯..+(√an-√(a(n+1)))/(an-a(n+1) )
= (√a1-√a2)/(-d )+(√a2-√a3)/(-d )+(√a3-√a4)/(-d )+⋯..+(√an-√(a(n+1)))/(-d )
= (√a1-√an)/(-d )
= (√a1-√(a(n+1)))/(-d )×(√a1+√(a(n+1)))/(√a1+√(a(n+1)))
= 1/(√a1+√an)× (a(n+1) –a1)/d ——1)
Now as we know a(n+1) = a1 + nd
So a(n+1) – a1= nd
Putting this value of a(n+1) – a1in eq 1)
We get required sum = n/(√a1+√(a(n+1)))

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