Question

Answer - :
t₁₀=24
In an A. P
=>S₁₀=126
Where,
a= 1st term
l= last term
Therfore,
We get,
=>10/2(a+t₁₀)=150
=>a+t₁₀=150/5
=>a+t₁₀=30 .......(1)
=>9/2(a+t₉)=126
=>a+t₉=126/9×2
=>a+t₉=14×2
=>a+t₉=28....... (2)
We subtracted equation (1) and( 2)
=>t₁₀ - t₉=2
common difference =d
=>d=2
=>10/2(2a+2(10−1))=150
=>2a+18=30
=>2a=30−18=12
=>a= 12/2
=>a=6

Answer 1

Answer:

it's right

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Step-by-step explanation:

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