Question

1+3+5+.................+101

Answer 1

Given :-

Arithmetic progession (AP)

◆ 1+3+5+..........+101

To find :-

The sum of given progession

• First find the value of n

• Now according to the formula

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

We have

a= 1

d= 3-1= 2

$\sf a_n= 101$

$\implies\sf 101= 1+(n-1)2\\ \\ \implies\sf 101-1= (n-1)2\\ \\ \implies\sf \cancel{\dfrac{100}{2}}= n-1\\ \\ \implies\sf 50+1= n\\ \\ \implies\sf n= 51$

• So number of terms in the given progession is 51

• Now find their sum

$\boxed{\sf{\dag\ \ S_n =\dfrac{n}{2}[a+a_n]}}$

Where ,

S= sum of progession

$\sf \ a_n$ = last term

a= first term

n= number of terms

d= common difference

By putting values in the formula

$\implies\sf S_{51}= \dfrac{51}{2}[1+101]\\ \\ \implies\sf S_{51}= \dfrac{51}{\cancel{2}}\times \cancel{102}\\ \\ \implies\sf S_{51}= 51\times 51\\ \\ \implies\sf S_{51}= 2601$

$\boxed{\sf{\dag\ \ Sum\ of \ given \ terms = 2601}}$

Answer 2

GivEn:-

A.P. :- 1+3+5+......+101

To find:-

• Sum of given terms

• Value of n

We know that,

$\dag \; \large\bold{\underline{\underline{\boxed{\sf{\purple{ \sf a_n = a + (n - 1)d}}}}}}$

From the given AP,

we find a and d :-

• First term (a) = 1

• Common difference (d) = 3 - 1 = 2

• $\sf a_n = 101$

★ Putting values in formula:-

$\implies \sf{ 101 = 1 + (n - 1)2}$

$\implies \sf{ 101 = 1 + 2n - 2}$

$\implies \sf{ 101 = 2n -1}$

$\implies \sf{ 101 + 1 = 2n}$

$\implies \sf{ \cancel{ \dfrac{102}{2}} = n}$

$\implies \sf{n = 51}$

• Number of term in the given progress is 51.

★ Finding the sum of AP.

We know that,

$\dag \; \large\bold{\underline{\underline{\boxed{\sf{\purple{ \sf S_n = \dfrac{n}{2} \bigg( a + \sf a_n \bigg) }}}}}}$

Here,

• $\sf a_n$ = last term of AP

• a = first term of AP

• n = Number of terms in an AP

• d = common difference between the numbers

★ Finding the sum of given progress:-

★ Putting values in formula:-

$\implies\sf{ \sf S_{51} = \dfrac{51}{2} \bigg( 1 + 101 \bigg)}$

$\implies\sf{ \sf S_{51} = \dfrac{51}{ \cancel{2}} × 102}$

$\implies\sf{ \sf S_{51} = 51 × 51}$

$\implies \sf{ \sf S_{51} = 2601}$

$\dag \; \large\bold{\underline{\underline{\boxed{\sf{\purple{ \sf S_n = 2601}}}}}}$

$\underline {\rule{261}{2}}$