Question

Find five numbers in AP whose sum is 12.5 and ratio of first and last is 2:3

Answer 1

Step-by-step explanation:

a+a+d+a+2d+a+3d+a+4d = 12.5

5a + 10d = 12.5

a + 2d = 2.5

यानी A(3) = 2.5

a/a+4d = 2/3

3a = 2a+8d

a = 8d

8d + 2d = 2.5

10d = 2.5

d = 0.25

a = 8*0.25 = 2

The five terms of this A.P.

2 , 2.25 , 2.50 , 2.75 , 3

Answer 2

Five term of A.P series are 2, 2.25, 2.5 , 2.75 ,3

Step-by-step explanation:

• Given data

        Sum of first five term = 12.5

        Ratio of first and fifth term are 2:3

        Means first term (a) = 2 k

           Fifth term     = 3 k = Last term

• From formula of sum of n term are

        $\mathbf{S_{n}=\frac{n}{2}\times (First\ term+ Last \ term)}$

• On putting respective value in above equation

        $\mathbf{12.5=\frac{5}{2}\times (2k+ 3k)}$

        $\mathbf{25=5\times 5k}$

        On solving, we get

        k=1

• So

        First term (a) =2×k=2×1=2

        Fifth term  =3×k=3×1=3

• We also know formula of $n^{th}$ term of A.P series

        $\mathbf{T_{n}=a+(n-1)d}$

        $\mathbf{3=2+(5-1)d}$

        $\mathbf{1=4d}$

        Means common difference (d) = 0.25

• Then

        First term = a = 2

        Second term = a+d =2 +0.25 =2.25

        Third term = a + 2d = 2+ 2×0.25=2.5

        Fourth term = a + 3d = 2+ 3×0.25=2.75

        Fifth term = a + 4d = 2+ 4×0.25=3