The first three terms of a geometric sequence are `x`, `y`,`z` and these have the sum equal to `42`. If the middle term `y` is multiplied by `5//4`, t

Question

The first three terms of a geometric sequence are `x`, `y`,`z` and these have the sum equal to `42`. If the middle term `y` is multiplied by `5//4`, t

The first three terms of a geometric sequence are `x`, `y`,`z` and these have the sum equal to `42`. If the middle term `y` is multiplied by `5//4`, the numbers `x`, `(5y)/(4)`, `z` now form an arithmetic sequence. The largest possible value of `x` is
A. `6`
B. `12`
C. `24`
D. `20`

Monjur Alahi

4 views

2025-01-04 04:42

1 Answers

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Salim Ahmed Kawsar · Accepted Answer · 2023-02-05T15:29:48+06:00

Correct Answer - C
`(c )` The three terms of the geometric sequence with the common ratio `r` are `x`, `xr`, `xr^(2)`. ,brgt `:.x+xr+xr^(2)=42`
After multiplying the middle term by `5//4`, we will get an arithmetic sequence. ,brgt `:.(5)/(4)xr-x=xr^(2)-(5)/(4)xr`
`:.2r^(2)-5r+2=0 :. r=(1)/(2)` and `r=2`
On substituting these in `x+xr+xr^(2)=42`, we get
`x=6` or `24`