We have:
\displaystyle a \frac {\left(\frac 3 4\right)^4 - 1} {\frac 3 4 - 1} = \frac {175} {64} a = 175
So:
\displaystyle a = 64
(b)
The sum to infinity is:
\displaystyle \frac {64} {1 - \frac 3 4} = 64 \times \frac 4 {4 - 3} = 64 \times 4 = 256
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The $9$th term in the sequence is:
\displaystyle 64 \times \left(\frac 3 4\right)^8 = \frac {6561} {1024}
The 10 th being:
\displaystyle 64 \times \left(\frac 3 9\right)^9 = \frac {19683} {1024}
We have:
\displaystyle \frac {19683} {1024} - \frac {6561} {1024} =-\frac {6561} {4096} = -1.6018\ldots = -1.602 \text { (3dp)}
So the absolute difference is 1.602 to 3 decimal places. (just -1.602 is fine)
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