Given:
a = sum of the first 20 terms of 4×51+5×61+6×71+....
b = sum of the first 22 terms of 3×41+4×51+5×61+6×71+....
Concept used:
xn = x1 + (n - 1) × d
Where xn = nth term
n = No. of term
x1 = 1st term
d = common difference
Calculation:
For the 1st series
a20 = 4 + (20 - 1) × 1
⇒ a20 = 4 + 19
⇒ a20 = 23
The last term will be 23×241,
Consider the first term only, i.e. 4×31,
We can write it as 31−41
After doing a similar operation in other terms, we will get
a = 41−51+51−61.........+231−241
⇒ a = 41−241
⇒ a = 246;−;1
⇒ a = 245
Again
b22 = 3 + (22 - 1) × 1
⇒ b22 = 3 + 21
⇒ b22 = 24
The last term will be 24×251,
Consider the first term only, i.e. 4×51,
We can write it as 41−51
After doing a similar operation in other terms, we will get
⇒ b = 31−41+41−51.........+241−251
⇒ b = 31−251
⇒ b = 7525;−;3
⇒ b = 7522
Now,
(ab)-1 = 1/(ab)
⇒ (ab)-1 = 1/(245 × 7522)
⇒ (ab)-1 = 1/(18011)
⇒ (ab)-1 = 11180
⇒ (ab)-1 = 16114
∴ Required value of (ab)-1 is 16114.