Question

a sum of 700 rupees is to be used to give seven cash prizes to student of a school for their overall academic performance if each prize is 20 less than its preceding prize find the value of each of the prizes ​

Answer 1

Answer

Given

• A sum of 700 rupees is to be used to give seven cash prizes to student of a school for their overall academic performance.

• Each prize is 20 less than its preceding prize.

To Find

The value of each of the prizes .

Solution

Now According To Question Let,

$\mathsf{\longrightarrow \: {1}^{st} \: price \: be (a_1) = a}$

$\mathsf{\longrightarrow \: {2}^{nd} \: price \: be (a_2) = a - 20}$

$\mathsf{\longrightarrow \: {3}^{rd} \: price \: be (a_3) = (a - 20) - 20 = a - 40}$

So on.....

Now we have formula to find Sum of A.P , that is...

$\mathsf{\longrightarrow \: Common \: difference (d) = a_2 - a_1 = (a - 20)- a = -20}$

$\mathsf{\longrightarrow \: Common \: difference (d) = a_3 - a_2= (a - 20)- a = -20}$

Since Difference are same we can say that it's in A.P

Now we know

$\boxed{ \mathsf{\bigstar \: S_n = \dfrac{n}{2}\bigg( 2a + (n - 1)d \bigg)}}$

Where

$\longrightarrow \: \mathsf{S_n = Sum \: of \: n \: terms}$

$\longrightarrow \: \mathsf{a = First \: term\: of \: AP}$

$\longrightarrow \: \mathsf{n = Number \: of \: terms }$

$\longrightarrow \: \mathsf{d = Common \: difference}$

Now Here,

$\longrightarrow \: \mathsf{S_n = 700rs}$

$\longrightarrow \: \mathsf{a = First \: price (a_1) = ? }$

$\longrightarrow \: \mathsf{d = -20 }$

$\longrightarrow \: \mathsf{n = Number \: of \: prices = 7}$

Putting this values in above formula we have

$\mathsf{\Longrightarrow \: S_n = \dfrac{n}{2} \bigg( 2a + (n - 1)d \bigg)}$

$\mathsf{\longrightarrow \: 700 = \dfrac{7}{2} \bigg( 2a + (7 - 1)(-20)\bigg)}$

$\mathsf{\longrightarrow \: 700 = \dfrac{7}{2} \bigg( 2a + (6)(-20)\bigg)}$

$\mathsf{\longrightarrow \: 700 = \dfrac{7}{2} \bigg( 2a -120 \bigg)}$

$\mathsf{\longrightarrow \: 700 = 7(a - 60)}$

$\mathsf{\longrightarrow \: 700 = 7a - 420}$

$\mathsf{\longrightarrow \: 700 + 420= 7a}$

$\mathsf{\longrightarrow \: 7a = 1120}$

$\mathsf{\longrightarrow \: a = \dfrac{1120}{7} = 160}$

Therefore we have,

$\mathsf{\longrightarrow \: {1}^{st} \: Price (a_1) = a = 160rs}$

$\mathsf{\longrightarrow \: {2}^{nd}\: Price (a_2) = a_1 - 20 \to 160rs - 20 = 140rs}$

$\mathsf{\longrightarrow \: {3}^{rd}\: Price (a_3) = a_2 - 20 \to 140rs - 20 = 120rs}$

$\mathsf{\longrightarrow \: {4}^{th}\: Price (a_4) = a_3 - 20 \to 120rs - 20 = 100rs}$

$\mathsf{\longrightarrow \: {5}^{th}\: Price (a_5) = a_4 - 20 \to 100rs - 20 = 80rs}$

$\mathsf{\longrightarrow \: {6}^{th}\: Price (a_6) = a_5 - 20 \to 80rs - 20 = 60rs}$

$\mathsf{\longrightarrow \: {7}^{th}\: Price (a_7) = a_6 - 20 \to 60rs - 20 = 40rs}$

$\rule{290}{1}$

$\large \underline{\underline{\red{ \mathsf{ \: \: \star \: More \: To \: Know \: \star \: \: }}}}$

$\mathsf{\longrightarrow \: {N}^{th} \: term \: of \: an \: A.P}$

$\boxed{\mathsf{\star \: a_n = a + (n - 1)d}}$

Where,

$\to \mathsf{a_n \: = {n}^{th} \: term}$

$\to \: \mathsf{a = First \: term\: of \: AP}$

$\to \: \mathsf{n = Number \: of \: terms }$

$\to \: \mathsf{d = Common \: difference}$

$\rule{290}{1}$

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