Question

the 6th term of an ap is 5 times the first term and the 11th term exceeds twice the 5th term by 3 find the 8th
term

Answer 1

hey dude ,

answer is --

let a is be its first term and d is the common difference
Given , a6 = 5a ......(1)
& a11 = 2(a5)+3 ......(2)

considered equation (1) , we get
a6 = 5a
=> a+ 5d = 5a
=> a - 5a = -5d
=> -4a= -5d
=> -4a+5d=0 ......(3)
now , considered equation (2)

a11 = 2(5a) + 3
=> a+ 10d = 2(a+4d)+3
=>a+10d = 2a + 8d +3
=> -a + 2d = 3 ... .. (4)

multiply equation (3)by 2 and (4) by 5
we get ,

-8a + 10d = 0 ....(5)
& -5a + 10d = 15 ....(6)
Now, subtract equation (5) from (6)
we get ,
-3a = -15
=> a = 5
put this value in equation (5) we get
-8 × 5 + 10d = 0
=> d = 4
so, a = 5 & d= 4

now ,

a8 = a + 7d

=> a8 = 5+ 7 ×4
= 5+28 = 33

hence , a8 = 33

★ hope it help you ★

Answer 2

The 8th term is 33.

Given:

$\bf a_6 = 5a ......(1) \\\\a_1_1 = 2(a_5) + 3 ......(2)$

Let ‘a’ be its first term, and ‘d’ is a common difference.

To find: 8th term

Solution:

Step 1: Simplify eq.1 by using the formula , nth term , $\bf a_n = a + (n - 1)d$:

$\bf a_6 = 5a \\\\a + (6 - 1)d = 5a \\\\a + 5d = 5a \\\\ a - 5a = - 5d \\\\-4a= - 5d \\\\-4a + 5d = 0 .... (3)$

Step 2: Simplify eq.2 by using the formula , nth term , $\bf a_n = a + (n -1)d$:

$\bf a_1_1 = 2(a_5) + 3 \\\\a + (11 -1)d = 2(a + (5 -1)d) + 3 \\\\a + 10d = 2(a + 4d) + 3 \\\\ a + 10d = 2a + 8d + 3 \\\\a - 2a + 10d - 8d = 3 \\\\-a + 2d = 3 .... (4)$

Step 3: Multiply eq. (3) by 2 and (4) by 5 and find a:

$\bf -8a + 10d = 0 ….(5) \\\\-5a + 10d = 15 ....(6) \\\\Subtract \ eq. (5) from (6) \\\\-5a + 10d - (-8a + 10d ) = 15 - 0 \\\\-5a + 10d + 8a - 10d = 15 \\\\-5a + 8a + 10d - 10d = 15 \\\\-3a = -15 \\\\a = \frac{15}{3} \\\\a = 5$

Step 4: Substitute this value of a in eq. (5) we get

$\bf -8a + 10d = 0 \\\\-8 \times 5 + 10d = 0 \\\\-40 + 10d = 0 \\\\10d = 0 + 40 \\\\10d = 40 \\\\d = \frac{40}{10} \\\\d = 4$

Step 5: Find the 8th term by using the formula , nth term ,$\bf a_n = a + (n -1)d$:

$\bf a_8 = a + 7d \\\\a_8 = 5 + 7 \times 4 \\\[a = 5 \ and \ d = 4] \\\\ a_8 = 5 + 28 \\\\a_8 = 33$

Hence the 8th term is 33.

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