Question

the sumof the 8th and the 13th term of an ap is 44 if its 10th term is three times its third term find the sum of 20 terms of the AP​

Answer 1

Given:-

Sum of 8th and 13th term of an A.P. is 44

To Find:-

Sum of 20 terms of the A.P.

From the question:-

We know,

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

8th term,

$\sf{a_8 = a+(8-1)d}$

= $\sf{a_8 = a+7d}$

13th term,

$\sf{a_{13} = a + (13-1)d}$

= $\sf{a_{13} = a + 12d}$

3rd term,

$\sf{a_3 = a+(3-1)d}$

= $\sf{a_3 = a+2d}$

10th term,

$\sf{a_{10} = a+(10-1)d}$

= $\sf{a_{10} = a+9d}$

Solution:-

8th term + 13th term = 44

= $\sf{(a+7d)+(a+12d) = 44}$

= $\sf{a+7d+a+12d = 44}$

= $\sf{2a + 19d = 44 \longrightarrow [i]}$

Now,

10th term is three times 3rd term

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

= $\sf{3a+6d = a + 9d}$

= $\sf{3a+6d-a-9d = 0}$

= $\sf{2a-3d = 0\longrightarrow[ii]}$

Subtracting eq.[ii] from eq.[i]

= $\sf{(2a+19d)-(2a-3d) = 44-0}$

= $\sf{2a+19d-2a+3d = 44}$

= $\sf{22d = 44}$

= $\sf{d = \dfrac{44}{22}}$

= $\sf{d = 2}$

Putting the value of d in eq.[i]

$\sf{2a + 19\times2 = 44}$

= $\sf{2a + 19\times 2= 44}$

= $\sf{2a + 38 = 44}$

= $\sf{2a = 44-38}$

= $\sf{a = \dfrac{6}{2}}$

= $\sf{a = 3}$

Verification:-

8th term + 13th term = 44

LHS

$\sf{a+7d + a + 12d}$

= $\sf{3 + 7\times2 + 3 + 12\times 2}$

= $\sf{6 + 14 +24}$

= $\sf{20+24}$

= $\sf{44 = (RHS)\:\:[Verified]}$

Now,

Sum of 20 terms,

We know,

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

= $\sf{s_{20} = \dfrac{20}{2}[2\times3 + (20-1)\times2]}$

= $\sf{s_{20} = 10[6+19\times2]}$

= $\sf{s_{20} = 10[6+38]}$

= $\sf{s_{20} = 10\times44}$

= $\sf{s_{20} = 440}$

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Formulas:-

• $\sf{a_n = a +(n-1)d]}$

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

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