Question

If the nth term of A.P be p, show that the sum of first (2n -1)terms of the A.P is (2n -1)p​

Answer 1

Step-by-step explanation:

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Answer 2

Step-by-step explanation :

ARITHMETIC PROGRESSION :-

• It is the sequence of numbers such that the difference between any two successive numbers is constant.

• In AP,

     a - first term

     d - common difference

     aₙ - nth term

     Sₙ - sum of n terms

• General form of AP,

        a , a+d , a+2d , a+3d , ..........

• Formulae :-

         nth term of AP,

         $\boxed{\bf a_n=a+(n-1)d}$

       Sum of n terms in AP,

          $\boxed{\bf S_n=\frac{n}{2}[2a+(n-1)d]}$

____________________________

Given,

• nth term = p

we know,

 aₙ = a+ (n - 1)d

  p = a + (n - 1)d

• Sum of first (2n - 1) terms

    $\bf S_n=\frac{n}{2}[2a+(n-1)d]$

 

In this formula, substitute n = (2n - 1)

           $S_{2n-1}=\frac{2n-1}{2}[2a+((2n-1)-1)d] \\\\\\ S_{2n-1}=\frac{2n-1}{2}[2a+(2n-2)d] \\\\\\ S_{2n-1} =\frac{2n-1}{2}[2(a)+2(n-1)d] \\\\\\ S_{2n-1}=\frac{2n-1}{2}[2(a+(n-1)d} \\\\\\S_{2n-1} =\frac{2n-1}{2}[2p] \\\\\\ \bf S_{2n-1}=(2n-1)p$

Hence proved!!