Question

The 13th term of an ap is 4 times its 3rd term. If its 5th term is 16, find the sum of its first 10 terms.

Answer 1

an = a1 + (n - 1)d

Find the 13th term:

a13= a + 12d

Find the 3rd term:

a3 = a + 2d

Given that 13th term is 4 times its 3rd term:

a + 12d = 4(a + 2d)

a + 12d = 4a + 8d

3a = 4d ----------------------- [ 1 ]

Given that the 5th term is 16:

16 = a +4d

a = 16 - 4d ----------------------- [ 2 ]

Sub { 2 } into [ 1 ]:

3(16 - 4d) = 4d

48 - 12d = 4d

16d = 48

d = 3  ----------------------- sub into equation [ 2 ]

Find a:

a = 16 - 4d

a = 16 - 4(3)

a = 4

Find the nth term:

an = a1 + (n - 1)d

an = 4 + (n - 1)3

an = 4 + 3n - 3

an = 1 + 3n

Find the 10th term:

a10 = 1 + 3(10)

a10 = 31

Find the sum of the first 10 terms:

Sn = n/2 (a1 + an)

S10 = 10/2 (4 + 31)

S10 = 175

Answer: The sum of the first 10 terms is 175

Answer 2

Answer:

$\text{Given} : a_{13} = 4 a_{3} \\\\a_5 = 16 \\\\\text{To find} : S_{10} \\\\\text{According to the question,}$

$\implies a_{13} = a + 12d \\\\\implies a_3 = a + 2d \\\\\implies a+ 12d = 4 ( a + 2d )\\\\\implies a + 12d = 4a + 8d \\\\\implies 4a - a = 12d - 8d \\\\\implies 3a = 4d \\\\\implies a = \dfrac{4d}{3}$

$\implies a_5 = a + 4d \\\\\implies a + 4d = 16\\\\\text{Substituting the value of a we get,} \\\\\implies \dfrac{4d}{3} + 4d = 16\\\\\implies \dfrac{4d + 12d}{3} = 16\\\\\implies 16d = 16 \times 3\\\\\implies 16d = 48$

$\implies d = \dfrac{48}{16} = 3 \\\\\implies a = \dfrac{4d}{3} = \dfrac{ 4 \times 3 }{3} = 4$

$S_{10} = \dfrac{n}{2} \: [ \:2a + ( n - 1 ) \:d \:]\\\\\implies S_{10} = \dfrac{10}{2} \: [ \: 2 ( 4 ) + ( 10 - 1 ) \: 3 \: ]\\\\\implies S_{10} = 5 \: [ 8 + 9 ( 3 ) ] \\\\\implies S_{10} = 5 \: [ 8 + 27 ] \implies 5 \: [ 35 ] \implies 175$

Hence the sum of first 10 terms is 175.