Question

Sum of 4 terms of an appointment is 10 sum of their squares is 30 find the ap

Answer 1

$\large\bf\underline{Given:-}$

• Sum of 4 terms = 10
• Sum of squares of terms = 30

$\large\bf\underline {To \: find:-}$

• AP

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

Let the 4 terms be ,a , a+ d, a+2d ,a+3d

$\underbrace{ \bf \dag \: According \: to \: question}$

• Sum of 4 terms = 10

$\dashrightarrow\rm\:a + ( a + d) + (a + 2d) + (a + 3d) = 10 \\ \\ \dashrightarrow\rm\:4a + 6d = 10 \\ \\ \rm \blacktriangleright \: divide \: both \: side \: by \: 2 \\ \\ \dashrightarrow\rm\:2a + 3d = 5.........(i)$

• Sum of their squares is 30

$\dashrightarrow\rm\: {a}^{2} + (a + d) {}^{2} + (a + 2d {)}^{2} + (a + 3d {)}^{2} = 30 \\ \\ \dashrightarrow\rm\: {a}^{2} + {a}^{2} + {d}^{2} + 2ad + {a}^{2} + 4 {d}^{2} + 4ad + {a}^{2} + {9d}^{2} + 6ad = 30 \\ \\ \dashrightarrow\rm\:4 {a}^{2} + 14{d}^{2} + 12ad = 30 \\ \\ \rm \blacktriangleright \: divide \: both \: side \: by \: 2 \\ \\ \dashrightarrow\rm\:2a {}^{2} + 7 {d}^{2} + 6ad = 15........(ii)$

$\rm \blacktriangleright \: squaring \: on \: both \: side \: of \: eq.(i)$

$\leadsto\rm\:(2a + 3d {)}^{2} = {5}^{2} \\ \\ \leadsto\rm\:4 {a}^{2} + 9 {d}^{2} + 12ad = 25........(iii)$

$\rm \dag \: multiplying \: eq.(ii) \: by \: 2 \: \\ \\ \rm \mapsto \: ({2a}^{2} + 7 {d}^{2} + 6ad = 15 )\times 2 \\ \\ \rm \mapsto \:4 {a}^{2} + 14 {d}^{2} + 12ad = 30.......(iv)$

Solving eq.(iii)and (iv).

$\rm\cancel{4 {a}^{2}} + 9 {d}^{2} + \cancel{12ad} = 25 \\ \rm \cancel{4 {a}^{2}} + 14 {d}^{2} + \cancel{ 12ad} = 30 \\ - \: \: \: \: \: \: \: \: \: - \: \: \: \: \: \: \: \: \: \: \: \: \: \: - \: \: \: \: \: \: \: \: - \\ \underline{ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: } \\ \: \: \: \: \: \rm \: \: \: \: \: \: \: - 5d {}^{2} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = - 5 \\ \rm \: \: \: \: \: \: \: \: \: \: \: \: \: \: d \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 1$

Putting value of d = 1 in eq.(i) ,

$\leadsto\rm \: 2a + 3d = 5 \\ \\ \leadsto\rm \: 2a + 3 = 5 \\ \\\leadsto\rm \: 2a = 2 \\ \\ \leadsto\rm \: a = 1$

AP will be :-

First term of AP = 1

2nd term of AP a+d = 2

3rd term of AP a+2d = 3

• AP = 1, 2 , 3....