Question

20 A boy find the 4 and 10' terms of an AP are 13 and 25 respectively.
a) Find the first term
b)find its common difference
c) Find its 17 term
d) Find the sum of its 10 term​

Answer 1

Step-by-step explanation:

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Answer 2

Answer:

Given:-

• 4th term in AP is 13
• 10th term in AP is 25

Find:-

• Common difference?
• First term?
• 17th term?
• Sum of its 10 terms

Solution:-

Common difference:-

From question, 4th term = 13

we know that,

$\: \: \: \: \: \: \: { \boxed{ \sf{ a_{n} = a + (n - 1)d}}}$

${ \sf{So, \: a + 3d = 13......(1)}}$

10th term = 25

${ \sf{So, \: a + 9d = 25......(2)}}$

Now solve equation (1) and (2):-

$\to{ \sf{(a + 9d) - (a + 3d) = 25 - 13}}$

$\to \sf{a + 9d - a - 3d = 12}$

$\to{ \sf{6d = 12}}$

$\to{ \sf{d = \frac{12}{6} = 2}}$

${ \boxed{ \therefore{ \sf{ \pink{Common \: difference = 2}}}}}$

First term:-

Now substitute the value of d in equation (1)

$\mapsto{ \sf{a + 3d = 13}}$

$\mapsto{ \sf{a + 3 \times 2 = 13}}$

${ \mapsto{ \sf{a + 6 = 13}}}$

${ \mapsto{ \sf{a = 13 - 6 = 7}}}$

${ \boxed{ \therefore{ \sf{ \pink{First \: term = 7}}}}}$

17th term:-

${ \boxed{ \sf{ a_{17} = a + 16d}}}$

${ \to{ \sf{ a_{17} = 7 + 16 \times 2 }}}$

${ \to{ \sf{ a_{17} = 7 + 32 }}}$

$\to{ \sf{ a_{17} = 39}}$

${ \boxed{ \therefore{ \pink{ \sf{17 \: term = 39}}}}}$

Sum of 10 terms:-

We know that,

$\: \: \: \: \: \: \: { \boxed{ \sf{ s_{n} = \frac{n}{2}(2a + (n - 1)d)}}}$

${ \mapsto{ \sf{ s_{10} = \frac{10}{2}(2 \times 7 + (10 - 1)2) }}}$

${ \mapsto{ \sf{ s_{10} = 5(14 + (9)2) }}}$

${ \mapsto{ \sf{ s_{10} =5( 14 + 18)}}}$

${ \sf{ \mapsto{ s_{10} = 5 \times 32}}}$

${ \mapsto{ \sf{ s_{10} = 160 }}}$

${ \boxed{ \therefore{ \sf{ \pink{Sum \: of \: 10 \: Terms = 160}}}}}$

Step-by-step explanation:

Therefore,

• Common difference = 2
• First term = 7
• 17th term = 39
• Sum of 10 terms = 160