Question

The sum first 10 terms of an AP is 15 and the sum of next 10 terms of AP is 215.Find the AP

Answer 1

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

• Sum of first ten terms = 15
• Sum of next ten terms = 215

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

The A.P.

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

We know,

$\sf{S_n = \dfrac{n}{2}[2a+(n-1)d]}$

Sum of first 10 terms = 15

= $\sf{S_{10} = \dfrac{10}{2}[2a + (10-1)d]}$

$\sf{\implies 15 = 5[2a+9d]}$

$\sf{\implies \dfrac{15}{5} = 2a + 9d}$

$\sf{\implies 3 = 2a + 9d}$

$\sf{\implies 2a + 9d = 3 \longrightarrow (i)}$

Now,

Sum of next 10 terms = 215

Therefore,

Sum of first 20 terms = 215 + 15 = 230

$\sf{S_{20} = \dfrac{20}{2}[2a+(20-1)d]}$

$\sf{\implies 230 = 10[2a+19d]}$

$\sf{\implies \dfrac{230}{10} = 2a + 19d}$

$\sf{\implies 23 = 2a + 19d}$

$\sf{\implies 2a + 19d = 23\longrightarrow (ii)}$

From eq.(i) and eq. (ii),

$\sf{\:\:\:\:\:2a + 9d = 3}$

$\sf{\:\:\:\:\:2a + 19d = 23}$

$\sf{\underline{(-)\:\:\:\:(-)\:\:\:\:\:\:(-)}}$

$\sf{By\:Sub \:-10d = -20}$

$\sf{\implies d = \dfrac{-20}{-10}}$

$\sf{\implies d = 2}$

$\sf{\underline{Putting\:the\:value\:of\:d\:in\:eq.(i)}}$

$\sf{2a+9d = 3}$

$\sf{2a + 9\times2 = 3}$

$\sf{\implies 2a + 18 = 3}$

$\sf{\implies 2a = 3-18}$

$\sf{\implies 2a = -15}$

$\sf{\implies a = \dfrac{-15}{2}}$

$\sf{\implies a = -7.5}$

Now,

a = -7.5

d = 2

A.P.:-

1st term

We know,

$\sf{a_n = a+(n-1)d}$

$\sf{a_1 = -7.5 + (1-1)\times 2}$

$\sf{\implies a_1 = -7.5 + 0}$

$\sf{\implies a_1 = -7.5}$

2nd term,

$\sf{a_2 = -7.5 + (2-1)\times2}$

$\sf{\implies a_2 = -7.5 + 1\times2}$

$\sf{\implies a_2 = -7.5 + 2}$

$\sf{\implies a_2 = -5.5}$

3rd term,

$\sf{a_3 = -7.5 + (3-1)\times2}$

$\sf{\implies a_3 = -7.5 + 2\times2}$

$\sf{\implies a_3 = -7.5 + 4}$

$\sf{\implies a_3 = -3.5}$

Therefore,

A.P. = -7.5, -5.5, -3.5, .......