Question

AP :- 1+3+5+........+101

Find the sum of the given AP​

Answer 1

$\huge\bf\red{\underline{\underline{Given}}}\::$

• $\sf\gray{(AP) \ : \ 1+3+5+..........+101}$

$\huge\bf\green{\underline{\underline{To\:Find}}}\::$

• $\sf\gray{The \ sum \ of \ given \ progession}$

$\huge\bf\pink{\underline{\underline{Solution}}}\::$

$\sf\underline\orange{Firstly \ we \ should \ find \ the \ value \ of \ n }$

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

$\sf\green{We \ have }$

• $\sf\gray{a \ = \ 1 }$
• $\sf\gray{d \ = \ 3-1 \ = \ 2}$
• $\sf\gray{ a_n \ = \ 101}$

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

$\star\:\:\sf\underline\red{Now \ let's \ find \ their \ sum }$

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

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

$\star\:\:\boxed{\bf{\red{ Sum\ of \ given \ terms = 2601}}}$

Answer 2

Given :

• A.P : 1+3+5+.....+101

To Find :

• Sum Of The A.P.

Solution :

• a = 1
• d = (5-3)= (3-1 )= 2
• l = 101 Last Term

Value of (n) =

• l = a + (n-1) d

• 101 = 1 + (n-1)2

• 101-1 = 2n-2

• 100 + 2 = 2n

• n = 51

Formula Used :

S = n/2 ( a + l )

S = 51/2 ( 1+101 )

S = 51/2 × 102

S = 51×51

S = 2601

Hence, Sum of the A.P. is 2601 .

Hope it helps.

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