Question

prove that in any arithmetic sequence ,the sum of the 25th term and the 15th term is twice the 20th term

Answer 1

SOLUTION :-

Let,

First term = a

Common difference = d

We have to prove that,

  $\sf 25^{th}$ term + $\sf 15^{th}$ term = 2 × $\sf 20^{th}$ term

We know that,

 $\bf a_n=a+(n-1)d$

$\bullet\sf \ a_{25}=a+(25-1)d$

       $=\sf a+24d$

$\bullet\sf \ a_{15}=a+(15-1)d$

       $= \sf a+14d$

$\bullet\sf \ a_{20}=a+(20-1)d$

       $= \sf a+19d$

$\longrightarrow \sf L.H.S = a_{25}+a_{15}$

               $= \sf (a+24d)+(a+14d}\\\\= 2a+38d$

$\longrightarrow \sf R.H.S =2\times a_{20}$

                $= \sf 2\times (a+19d)\\\\=2\times a +2\times 19d \\\\= 2a+38d$

$\bf\therefore L.H.S = R.H.S$

Hence proved.

ADDITIONAL INFORMATION :-

• The nth term of AP, $\bf a_n=a+(n-1)d$

• Sum of n terms of AP, $\bf S_n = \dfrac{n}{2}\times (2a+(n-1)d$

Answer 2

✉TO RPOVE:

• 25th term + 15th term = 2(20th term)

✉SOLUTION:

Here,

Let the first term be a

common difference be d

now,

we know that,

$\tt{ ✑ a_n=a+(n-1)d}$

now,

$\tt{ ➟ a_{25}=a+(25-1)d}$

$\tt{ ➟ a_{25}=a+(24)d}$

$➟ \tt{a_{25} = a + 24d}$

now,

$\tt{ ➟ a_{15}=a+(15-1)d}$

$\tt{ ➟ a_{15}=a+(14)d}$

$\tt{ ➟ a_{15}=a+14d}$

now,

$\tt{ ➟ a_{20}=a+(20-1)d}$

$\tt{ ➟ a_{20}=a+(19)d}$

$\tt{ ➟ a_{20}=a+19d}$

now,

$\tt{ ➨ a_{25} +a_{15} = 2( a_{20} )}.....(i)$

where,

$\tt{ ➤ a_{25} = a + 24d}$

$\tt{ ➤ a_{15}=a+14d}$

$\tt{ ➤ a_{20}=a+19d}$

substituting these values in eq.(i) we got

$\tt{ ➲ a + 24d +a + 14d = 2( a + 19d)}$

solve this

$\tt{ ➥ 2a + 38d = 2a + 38d}$

here, L.H.S = R.H.S

Hence, proved