Question

The 11th term of an AP is 16 and the 21st term of an AP is 29.find the 16th term of this AP.

Answer 1

Answer:

a₁₁ = 16 or a + 10d = 16 ..(i)

a₂₁ = 29 or a + 20d = 29 ..(ii)

Solving Equation (i) and (ii), we get,

d=13/10

a=3

a₁₆ = a + (n-1) d

a₁₆ = 3 + (16-1)13/10

a₁₆ = 3+(15)13/10

a₁₆ = 3 + 39/2

a₁₆ = 6+39/2

a₁₆ = 45/2

a₁₆ = 22.5

Answer 2

Answer :

☞ 16th term of AP is 45/2.

Solution :

Given :

• The 11th term of an AP is 16 .
• The 21st term of an AP is 29.

To Find :

• The 16th term of AP.

It is given that,

• The 11th term of an AP is 16.

So,

$\red\bigstar\bf T_{11} = 16 \\ \implies \bf\pink{ a+10d=16}……(i) \\ \\ \implies\bf a = 16-10d ……(ii)$

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• The 21st term of an AP is 29.

So,

$\red\bigstar\bf T_{21} = 29 \\ \implies \bf\pink{ a+20d=29}……(iii)$

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Substitute (ii) in (iii)

$\\ \implies \bf{ a+20d=29}……(iii) \\\\\implies \sf (16-10d)+20d= 29 \\\\\implies \sf 16-10d+20d=29 \\\\\implies \sf 16+10d=29 \\\\\implies \sf 10d=29-16 \\\\\implies \sf 10d=13 \\\\\implies \sf \underline{\boxed{\blue{\mathfrak{d=\frac{13}{10}}}}}$

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Putting the value of d in (i) ....

$\sf a+10d=16 \\\\\implies \sf a+\cancel{10}(\frac{13}{\cancel{10}})=16 \\\\\implies \sf a + 13 = 16 \\\\\implies \sf a = 16-13\\\\\implies \sf \underline{\boxed{\blue{\mathfrak{a = 3}}}}$

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Now, we can easily find 16th term

we know that,

$\red\bigstar \boxed{T_n = a +(n-1)d }$

we have ,

• a = 3
• d = 13/10
• n = 16

Substituting the values,

$\sf T_{16} = 3+(16-1)d \\\\\implies\sf T_{16} =3+(^3 \cancel{15})\frac{13}{^2 \cancel{10}} \\\\\implies\sf T_{16} = 3 +(3)\times\dfrac{13}{2} \\\\\implies\sf T_{16} = 3 + \dfrac{39}{2} \\\\\implies\sf T_{16} = \dfrac{6+39}{2} \\\\\implies\sf \boxed{T_{16} = \dfrac{45}{2}}$

Hence,

16th term of AP is 45/2.

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