GO Classes CS/DA 2025 | Weekly Quiz 1 | Fundamental Course | Question: 4

Answer 1

We know that in an $\text{A.P.}$, the sum of the terms equidistant from the beginning and end is always the same and is equal to the sum of the first and last term.

Therefore, $d=b-a$

$\Rightarrow a_1+a_{24}=a_5+a{20}=a_{10}+a_{15}$

It is given that

$(a_1+a_{24})+(a_5+a_{20})+(a_{10}+a_{15})=225$

$\Rightarrow (a_1+a_24)+(a_1+a_{24})+(a+a_{24})=225$

$\Rightarrow 3(a_1+a_{24})=225$

$\Rightarrow a_1+a_{24}=75$

We know that

$\text{S}_n=\dfrac{n}{2}[a+l],$

Where $a$ is the first term, and $l$ is the last term of an $\text{A.P.}$

$\text{S}_{24}=\dfrac{24}{2}[a_1+a_{24}]=12\times 75=900.$

Answer 2

Given,

            $a_{1}+a_{5}+a_{10}+a_{15}+a_{20}+a_{24} = 225$ 

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We know that the nth term of an AP is given by,  $a_{n} = a_{1} + (n-1)d$

$\therefore a_{5} = a_{1} + 4d$ , $a_{10} = a_{1} + 9d$ , $a_{15} = a_{1} + 14d$ , and so on..

Putting these values in our given equation, we get

$a_{1} + (a_{1} + 4d) + (a_{1} + 9d) + (a_{1} + 14d) + (a_{1} + 19d) + (a_{1} + 23d) = 225$

$\Rightarrow 6a_{1} + 69d = 225$

$\Rightarrow 2a_{1} + 23d = 75$

We know that sum of n terms of an AP is,  

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

where a is the first term and d is the common difference.

$ \therefore S_{24} = \frac{\large 24}{\large 2}[2a_{1}+(24-1)d]$

$\Rightarrow S_{24} = 12[2a_{1}+23d]$

$\Rightarrow S_{24} = 12 \times 75 = 900.$ 

 

 

Comments

Nice bro,
You mix Modular arithmetic and A.P

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where is modular arithmetic here?

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Only AP concepts are used in this answer.

Answer 3

we know ${a}$$_{n}$=${a}$$_{1}$${+(n-1)d}$

so  ${a}$$_{24}$${=a}$$_{1}$${+23d}$

      ${a}$$_{20}$${=a}$$_{1}$${+19d}$

     ${a}$$_{5}$${=a}$$_{1}$${+4d}$

    now ${a}$$_{20}$${+a}$$_{5}$${=a}$$_{1}$${+4d+}$${a}$$_{1}$${+19d=2a}$$_{1}$${+23d}$${=a}$$_{24}$${+a}$$_{1}$

similarly  ${a}$$_{10}$${+a}$$_{15}$${=a}$$_{24}$${+a}$$_{1}$

${a}$$_{1}$${+a}$$_{5}$${+a}$$_{10}$${+a}$$_{15}$${+a}$$_{20}$${+a}$$_{24}$${=3a}$$_{24}$${+3a}$$_{1}$${=3(a}$$_{24}$${+a}$$_{1}$${)=225}$

hence ${a}$$_{1}$${+a}$$_{24}$${=225/3=75}$

now ${S}$$_{24}$${=24(a}$$_{1}$${+a}$$_{24}$${)/2=24*75/2=900}$

${Ans.  S}$$_{24}$${=900}$