Question

If 6th term of an APis 19 and its 17th term is 41 then its 50th term of AP.​

Answer 1

Answer :

$\large{\star\:\:\boxed{\bf{a_{50}\:=\:107}}\:\:\star}$

Explanation :

Given :–

• a₆ = 19
• a₁₇ = 41

To Find :–

• a₅₀ (50th term of this A.P.)

Formula Applied :–

• $\boxed{\bf{\star\:\:a_n\:=\:a\:+\:(n\:-\:1)d\:\:\star}}$

Solution :–

We have ,

$\rightarrow\sf{a_6\:=\:19}$

$\rightarrow\sf{a_6\:=\:a\:+\:(6\:-\:1)d}$

$\rightarrow\sf{19\:=\:a\:+\:5d\:\:-----}\bf{(1)}$

We also have ,

$\rightarrow\sf{a_{17}\:=\:41}$

$\rightarrow\sf{a_{17}\:=\:a\:+\:(17\:-\:1)d}$

$\rightarrow\sf{41\:=\:a\:+\:16d\:\:-----}\bf{(2)}$

Subtracting Equation(2) from Equation(1) :-

$\rightarrow\sf{41\:-\:19\:=\:a\:+\:16d\:-\:(a\:+\:5d)}$

$\rightarrow\sf{22\:=\:a\:+\:16d\:-\:a\:-\:5d}$

$\rightarrow\sf{11d\:=\:22}$

$\rightarrow\sf{d\:=\:\dfrac{22}{11} }$

$\rightarrow\bf{d\:=\:2 }$

Now putting this 'd' in Equation(1) :-

$\rightarrow\sf{19\:=\:a\:+\:5(2)}$

$\rightarrow\sf{19\:=\:a\:+\:10}$

$\rightarrow\sf{a\:=\:19\:-\:10}$

$\rightarrow\bf{a\:=\:9}$

Now , we have a = 9 , d = 2 and n = 50 .

Putting these values in the Formula :

$\rightarrow\sf{a_n\:=\:a\:+\:(n\:-\:1)d}$

$\rightarrow\sf{a_{50}\:=\:9\:+\:(50\:-\:1)(2)}$

$\rightarrow\sf{a_{50}\:=\:9\:+\:(49\:\times\:2)}$

$\rightarrow\sf{a_{50}\:=\:9\:+\:98}$

$\rightarrow\boxed{\bf{a_{50}\:=\:107}}$

∴ The 50th Term of this A.P. is 107 .

Answer 2

◙ Given ◙

$\bigstar\ \textsf{6th term of an AP is 19.}$

$\bigstar\ \textsf{17th term of the AP is 41.}$

◙ To Find ◙

$\blacksquare\ \sf{ The\ 50th\ term\ of\ the\ AP.}$

◙ Solution ◙

$\sf{We\ know:}\\\\\bullet\ \bf{a_6=a+5d=19}\:\:\:\: \cdots(i) \\\bullet\ \bf{a_{17}=a+16d=41}\:\:\: \cdots(ii)$

$\sl{Subtracting\ equation\ (i)\ from\ (ii):}$

$\tt{a+16d=41}\\\tt{-}\\\tt{a+5d=19}\\\rule{60}{1.5}\\\tt{11d=22}$

$\sl{Calculating\ further:}$

$\sf{\longmapsto d=\dfrac{22}{11}}\\\\\sf{\longmapsto d=2}$

$\sl{Putting\ the\ value\ of\ d\ in\ equation\ (i):}$

$\sf{a+5d=19}\\\\\sf{\longmapsto a+5(2)=19}\\\\\sf{\longmapsto a+10=19}\\\\\sf{\longmapsto a=19-10}\\\\\sf{\longmapsto a=9}$

___________________

$\sf{Now,}$

$\bf{a_{50}=a+49d}\\\\\sf{\longmapsto a_5_0=9+49(2)}\\\\\sf{\longmapsto a_5_0=9+98}\\\\\boxed{\boxed{\sf{\longmapsto a_5_0=107}}}$

$\bf{\underline{Hence,\ the\ 50th\ term\ of\ the\ AP\ is\ 107.}}$