Question

Please help me solve this question please​

Answer 1

Answer:-

The Four Consecutive Terms Of The A.P. Are -7, -2, 3, 8 respectively.

Explanation:-

Given:-

• Sum of 4 consecutive terms of an AP = 2.

• Sum of 3rd and 4th term = 11.

Fromula Used:-

• ↦ an = a + (n-1)d.

• ↦ Sn = n/2(a + an)

Where,

• an = nth term.
• Sn = Sum of first n terms.
• a = First term.
• n = Number of terms.
• d = Common Difference.

Solution:-

So Let,

• 1st term be a.
• 2nd term be a + d.
• 3rd term be a + 2d
• 4th term be a + 3d.

ATQ:-

↦ S4 = 2.

↦ n/2(a + a4)= 2.

↦ 4/2[a + (a + 3d)] = 2.

↦ 2 (2a + 3d) = 2.

↦ 2a + 3d = 2/2.

↦ 2a + 3d = 1. -----eq(1)

Also,

↦ a3 + a4 = 11.

↦ (a + 2d) + (a+3d) = 11.

↦ 2a + 5d = 11. -----eq(2)

By Subtracting eq(1) from eq(2) we get,

↦ (2a + 5d) - (2a+3d) = 11 - 1.

↦ 2a + 5d - 2a - 3d = 10.

↦ 2d = 10.

↦ d = 10/2.

↦ d = 5.

By Putting The Value Of d in eq(1) we get,

↦ 2a + 3d = 1.

↦ 2a + 3(5) = 1.

↦ 2a + 15 = 1.

↦ 2a = 1 - 15.

↦ 2a = -14.

↦ a = -14/2.

↦ a = -7.

Therefore,

↦ First term = a = -7.

↦ Second Term = a + d = -2.

↦ Third Term = a + 2d = 3.

↦ Fourth Term = a + 3d = 8.

So The Required AP Is -7, -2, 3, 8,........

Answer 2

Heya !

Given:-

• Sum of four consecutive terms = 2
• Sum of 3rd and 4th term = 11

To find :-

• The terms of the A.P.

Formulas to be used:-

$\huge \fbox \blue{S=n/2(a+an)}$

$\huge \fbox \blue{an=a+(n-1)d}$

Solution:-

S4 = n/2(a+a4)

2 = 4/2 (a+a4)

1 = a + 4a

1 = a + (a+3d)

1 = 2a + 3d ...............eq.1

acc. to 2nd condition:-

a3 + a4 = 11

a+2d + a+3d = 11

2a+5d = 11 .................. eq.2

subtracting eq. 1 from eq.2

2a+5d=11

2a+3d=1

- - -

2d = 10

d = 5

From eq. 1

2a+5(5)=11

2a+25 = 11

2a = -14

a = -14/2

a = -7

therefore,

a2 = a+d = -7+5 = -2

a3 = a+2d = -7+2(5) = 3

a4 = a+3d = -7+3(5) = 8.

Hence the a.p is -7,-2,3,8