Question :

Let a₁, a₂ …………….. a₂₀₂₄ be an Arithmetic Progressions Such that a₁ + ( a₅ + a₁₀ + a₁₅ +………….. + a₂₀₂₀ ) + a₂₀₂₄ = 2233 Then a₁ + a₂ + a₃ + …………….+ a₂₀₂₄ is

Sequence and series JEE Main 2025 Shift – II with solution

Solution :

Find the sum of a₁ + a₂ + a₃ + …………….+ a₂₀₂₄

By using S_n Formula

S_n = $\frac{n}{2}$ [ a + l ]

S₂₀₂₄ = $\frac{2024}{2}$ [ a₁ + a₂₀₂₄ ]

= 1012 [ a₁ + a₂₀₂₄ ]

Given

a₁ + ( a₅ + a₁₀ + a₁₅ +………….. + a₂₀₂₀ ) + a₂₀₂₄ = 2233

By using Property :

Sum of the terms equidistant from begining and end is same i.e for an A.P a₁, a₂, a₃, a₄…………… a_n we have a₁ + a_n = a₂ + a_{n -1} = a₃ + a_{n – 2} = …………..

So, a₁ + a₂₀₂₄ = a₅ + a₂₀₂₀ = a₁₀ + a₂₀₁₅ = ……

Total number of terms = $\frac{2020}{5}$ = 404

= 2 ( 202)

Number of pairs = 202

( a₁ + a₂₀₂₄ ) + ( a₅ + a₂₀₂₀ ) + ( a₁₀ + a₂₀₁₅ ) + ………… = 2233

203 ( a₁ + a₂₀₂₄ ) = 2233

a₁ + a₂₀₂₄ = 11

S₂₀₂₄ = 1012 [ a₁ + a₂₀₂₄ ]

S₂₀₂₄ = 1012 ( 11)

S₂₀₂₄ = 11132

Practice Questions :

1. Find the terms of an A.P where sum of the 3 terms is 27 and product is 81 ?

Solution :

Given

Three terms as a – d, a , a + d are in A.P

a – d + a + a + d = 27

3a = 27

a = 9

Product of three terms is ( a – d ) (a) ( a + d) = 81

( a² – d²) ( 9) = 81

a² – d² = 9

81 – d² = 9

d² = 81 – 9

d² = 72

d = $\pm\sqrt{72}$

d = $\pm6\sqrt2$

The three terms are a – d, a , a + d

• d = + $6\sqrt{2}$ $\rightarrow$ terms are 9 – 6$\sqrt{2}$, 9, 9 + 6$\sqrt{2}$

• d = – 6$\sqrt{2}$ $\rightarrow$ terms are 9 + 6$\sqrt{2}$, 9 , 9 – 6$\sqrt{2}$

2. In four terms of A.P If sum of four terms is 4 and product is 0 then find all 4 terms ?

Solution :

The four terms are a – 3d, a – d, a + d , a + 3d

Sum of four terms = a – 3d + a – d + a + d + a + 3d = 4

4a = 4

a = 1

Product of four terms are ( a – 3d) ( a – d )( a + d) (a + 3d) = 0

( 1 – 3d) ( 1 – d) ( 1 + d) ( 1 + 3d) = 0

( 1 – d²) ( 1 – 9d²) = 0

assume d² = t

( 1 – t )( 1 – 9t) = 0

1 – 10t + 9t² = 0

9t² – 10t + 1 = 0

9t² – 9t -t +1 = 0

9t ( t -1) -1( t – 1) = 0

t = 1, $\frac{1}{9}$

d² = 1, $\frac{1}{3}$

d = $\pm$ 1, $\pm$ $\frac{1}{3}$

• d = $\frac{1}{3}$ the four terms are 0, $\frac{2}{3}$, $\frac{4}{3}$, 2
• d = 1 the four terms are -2, 0, 2, 4

3. Insert 7 AMs in between $\frac{1}{2}$ and 3 then find 5^th AM is

Solution :

$\frac{1}{2}$ A₁, A₂, A₃, A₄, A₅, A₆, A₇, 3

d = $\frac{ last term – a }{ number of terms – 1}$

d = $\frac{ 3 – 1/2}{9 -1}$

d = $\frac{5/2}{8}$

d = $\frac{5}{16}$

A₅ = a + 5d

A₅ = $\frac{1}{2}$ + 5d

A₅ = $\frac{1}{2}$ + $\frac{5. 5}{16}$

A₅ = $\frac{1}{2}$ + $\frac{25}{16}$

A₅ = $\frac{33}{16}$

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