Question

In an A.P., if the 12thterm is -13 and the sum of its first four terms is 24, find the sum of its ten terms

Answer 1

Answer:

t12=a+11d=-13(since tn=a+(n-1)d

s4(sum of four terms) is=24

s10(sum of ten terms)we need to find

now let a+11d=-13.....(1)

given s4=24(Sn=n/2(2a+(n-1)d)

4/2(2a+(4-1)d)=24

2(2a+3d)=24

2a+3d=12....(2)

now solve(1) and (2)

after solving a(first term=9)

common difference(d)=-2

now s10(sum of ten terms)=10/2(2(9)+(10-1)(-2)

s10=0

so sum of ten terms is zero

Step-by-step explanation:

Answer 2

Aɴsᴡᴇʀ

given :-

• 12th term = -13
• sum of first 4 terms = 24

to find :-

• sum of first ten terms

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solution :-

• a = first term
• n = term number
• an = nth term
• d = common difference
• l = last term
• Sn = sum of first n terms

and we know that,

$\boxed{ \pink{ \bold{a_{n} = a + (n - 1)d}}} \: \: \: \: \: \: \: \: \: \\ \\ \boxed{ \pink{ \bold{S_{n} = \frac{n}{2} \{2a + (n - 1)d \}}}} \\ \\ \boxed{ \pink{ \bold{S_{n} = \frac{n}{2} (a + l)}}} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

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here,

• 12th term = -13
• n = 12

$\bold{ \implies a_{12} = a + (12 - 1)d} \: \: \: \: \: \: \: \: \: \: \: \\ \\ \bold{ \implies -13 = a + 11d } \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \bold{ \implies a + 11d + 13 = 0}........(i)$

and

• sum of first four term is 24
• n = 4

$\bold{ \implies S _{4} = \cancel{\frac{4}{2}} \{ 2a + (4 - 1)d\}} \: \: \: \: \: \: \: \\ \\ \bold{ \implies 24 = 2(2a + 3d)} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \bold{ \implies2a + 3d = \cancel{\frac{24}{2} }} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \bold{ \implies 2a + 3d = 12} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \bold{\implies2a + 3d - 12 = 0} .........(ii)$

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now, we have two equations.

• a + 11d + 13 = 0 .......(i)
• 2a + 3d - 12 = 0 .......(ii)

from equation (i)

a = -11d - 13 .....(iii)

now substitute this value of a in (ii)

$\bold{ \implies 2( - 11d - 13) + 3d - 12 = 0} \\ \\ \bold{ \implies - 22d - 26 + 3d - 12 = 0} \: \: \: \: \: \: \\ \\ \bold{\implies - 19d - 38 = 0}\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \bold{\implies - 19d = 38}\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \bold{\implies d = \cancel{ \frac{38}{ - 19}}} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \implies \boxed{\bold{ d = - 2 }}\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

now, substitute the value of d in eq.(iii)

$\bold{ \implies a = - 11( - 2) - 13 } \\ \\ \bold{ \implies a = 22 - 13} \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \implies \boxed{ \bold{ a = 9}} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

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now, we get

• a = 9
• d = -2

we have to find sum of first ten terms

• n = 10

$\bold{ \implies S _{10} = \cancel{\frac{10}{2}} \{2 \times 9 + (10 - 1) - 2 \} } \\ \\ \bold{ \implies S_{10} = 5 \{18 + (9 \times - 2)\} } \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \bold{ \implies S _{10} = 5(18 - 18)} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \bold{ \implies S_{10} = 5 \times 0} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \implies\boxed{ \red{\bold{ S_{10} = 0 }}} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

so, sum of first ten terms of this AP is 0