Question

In An A.P. ,
Given :- T₁₀ = 10T₁
Prove That :- T₆ = 4T₄​

Answer 1

$\text{I think your question should be:}$

$\text{In an A.P, Given: T_{10}=10\,T_1 Prove that: T_6=3\,T_2 }$

$\textbf{Given:}$

$\text{In an A.P,}\;\;T_{10}=10\;T_1$

$\textbf{To prove:}$

$T_6=3\;T_2$

$\textbf{Solution:}$

$\text{We know that,}$

$\text{The n th term of the A.P}\,a,\,a+d,\,a+2d,..........\;\text{is}$

$\boxed{\bf\,T_n=a+(n-1)d}$

$\text{Consider,}$

$T_{10}=10\;T_1$

$\implies\,a+9d=10\,a$

$\implies\,9d=10\,a-a$

$\implies\,9\,d=9\,a$

$\implies\boxed{\bf\,d=a}$

$\text{Now,}$

$T_6$

$=a+5d$

$=a+5a$

$=6a$.........(1)

$3\;T_2$

$=3(a+d)$

$=3(a+a)$

$=3(2a)$

$=6a$.........(2)

$\text{From (1) and (2), we get}$

$\boxed{\bf\,T_6=3\;T_2}$

$\textbf{Find more:}$

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