Question

The sum of first 20 terms of an ap is 400 and sum of first 40 terms is 1600. Find the sum of its 10terms

Answer 1

$\Large{\textbf{\underline{\underline{According\;to\;the\;Question}}}}$

Assume

First term be a

Also,

Common difference be d

Now,

Sum of the 1st 20 terms = S(20)

S(20) = 20/2(2a + 19d)

Putting the values,

400 = 20/2(2a + 19d)

400 = 10(2a + 19d)

2a + 19d = 40 ............................. (1)

Here,

S(40) = 40/2(2a + 39d)

Putting the values,

1600 = 20(2a + 39d)

2a + 39d = 80 ............................ (2)

Now,

From (1) & (2) we get,

20d = 40

d = 40/20

d = 2

Now,

Substitute the value of d in (1),

2a + 19d = 40

2a + 19(2) = 40

2a + 38 = 40

2a = 40 - 38

2a = 2

a = 2/2

a = 1

Now,

S(10) = [2 × 1 + (10 - 1)2]

S(10) = 5[2 + 9 × 2]

S(10) = 5[2 + 18]

S(10) = 5 × 20

S(10) = 100

Therefore,

Sum of its 1st 10 terms = 100

Answer 2

Let us assume that ,

The first term and common difference of an AP be " a " and " d "

First Condition :

The sum of first 20 terms of an ap is 400

Thus ,

$\sf \Rightarrow 400 = \frac{20}{2} (2a + 19d)</p><p> \\ \\ \sf \Rightarrow </p><p>400 = 10(2a + 19d)</p><p> \\ \\ \sf \Rightarrow </p><p>2a + 19d = 40 \: - - \: (i)$

Second Condition :

The sum of first 40 terms of an AP is 1600

Thus ,

$\sf \Rightarrow 1600 = \frac{40}{2}(2a + 39d) \\ \\ \sf \Rightarrow 1600 = 20(2a + 39d) \\ \\ \sf \Rightarrow </p><p></p><p>2a + 39d = 80 \: - - \: (ii)$

Subtract eq (i) from eq (ii) , we get

$\sf \Rightarrow 20d = 40</p><p> \\ \\ \sf \Rightarrow </p><p>d = \frac{40}{20} </p><p> \\ \\ \sf \Rightarrow </p><p>d = 2$

Put the value of d = 2 in eq (i) , we get

$\sf \Rightarrow 2a + 19(2) = 40 \\ \\ \sf \Rightarrow </p><p></p><p>2a + 38 = 40 \\ \\ \sf \Rightarrow </p><p></p><p>2a = 2 \\ \\ \sf \Rightarrow </p><p></p><p>a = 1$

Now , we have to find the sum of first 10 terms of an AP

Thus ,

$\sf \Rightarrow S_{10}= \frac{10}{2} \{2 × 1 + (10 - 1)2 \}</p><p></p><p> \\ \\ \sf \Rightarrow S_{10}= 5 \{2 + 9 × 2 \}</p><p> \\ \\ \sf \Rightarrow </p><p>S_{10}= 5 × 20</p><p></p><p> \\ \\ \sf \Rightarrow S_{10}= 100</p><p>$

$\therefore \bold{ \underline{ \sf{The \: sum \: of \: first \: 10 \: terms \: of \: an \: AP \: is \: 100}}}$