Question

a person invests ₹4000 per month in recurring deposit for 3/2 years and gets ₹76560 on maturity.
find:the rate of interest and the interest received by him.​

Answer 1

Answer:

$A=(P×n)+P×2×12n(n+1)×100r</p><p></p><p>⟹4950=(400×12)+400×2×1212(13)×100r⟹26r=Rs.150⟹r=5.7%</p><p></p><p>$

Step-by-step explanation:

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Answer 2

Step-by-step explanation:

A⟹4950=(400×12)+400×2×1212(13)×100r⟹26r=Rs.150⟹r=5.7%A=(P×n)+P×2×12n(n+1)×100r⟹4950=(400×12)+400×2×1212(13)×100r⟹26r=Rs.150⟹r=5.7