Question

The 24th term of an AP is twice its 10th term. Show that its 72nd term is 4 times its 15th term.

Answer 1

Dear Student!!

Let a and d be the first term and common difference of the given A.P.  

Given:-

24th term of an AP is twice its 10th term.

Formula of Arithmetic Progression:- Tn=a+(n-1)d

T24=a+(24-1)d

T24=a+23d -----(A)

Again

T10=a+(10-1)d

T10=a+9d -------(B)

Considering on Question;-

T24= 2× T10

Substitute the value of Required terms;-

From Equation A and B , we gets

T24= 2× T10

=> a+23d=2 × a+9d

=> a+23d=2(a+9d)

=> a+23d=2a+18d

=>a-2a=18d-23d

=> -a=-5d

•°• a=5 d

Here, First term of Arithmetic Sequence or Progression = 5d ------(1)

According to the Question!-

Show that its 72nd term is 4 times its 15th term.

T72=a+(72-1)d

T72=a+71d ------(C)

Again;-

T15=a+(15-1)d

 T15=a+14d --------(D)

From Equation 'C' and 'D', We gets;-

Again, Considering on Question follow;-

T72=4×T15

a+71d=4×a+14d

=> a+71d=4(a+14d)

=> a+71d=4a+56d

Here, Using Equation 1 we gets,

a+71d=4a+56d

Substitute the value of a in Given Equation;-

a+71d=4a+56d

5d+71d=4(5d)+56d

5d+71d=20d+56d

76d=76d

Hence, It's Proved!!

Conclusion:- Yes, that its 72nd term is 4 times its 15th term.

Thank you!!!

Answer 2

Hey there !!


➡ Given :-

→ a $\tiny 24$ = 2 ( a $\tiny 10$ ).


➡ To prove :-

→ a $\tiny 72$ = 4 ( a $\tiny 15$ ).


➡ Solution :-


we have,

→ a $\tiny 24$ = 2 ( a $\tiny 10$ ).

=> a + 23d = 2( a + 9d ).

=> a + 23d = 2a + 18d.

=> 2a - a = 23d - 18d.

=> a = 5d.


we have, A/Q

a $\tiny 72$ = 4 ( a $\tiny 15$ ).

=> a + 71d = 4 ( a + 14d ).

[ Putting the value of ‘a’ ].

=> 5d + 71d = 4( 5d + 14d ).

=> 76d = 4 × 19d.

$\huge \boxed{ \boxed{ \bf => 76d = 76d. }}$


$\huge \boxed{ \boxed{ \bf{ \mathhbb{ LHS = RHS. }}}}$


✔✔ Hence, it is proved ✅✅.

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THANKS


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