Question

22. The first term of an A.P is 14 and the sums of the first five terms and the first ten terms are
equal is magnitude but opposite in sign. The 3rd term of the AP is​

Answer 1

Step-by-step explanation:

▶As we have given , the sums of the first five terms and the first ten terms are equal is magnitude but opposite in sign

therefor,

S5 = -(S10)

First term of an A.P is 14

a = 14

▶we know the formula to find out the sum of terms,

★ Sn = n / 2 ( 2a + ( n - 1 ) d)

∴ (S5) = -(S10)

→[(5/ 2 )( 2a + ( 5 - 1) d) ]

= [ ( 10 / 2 ) ( 2a + ( 10 - 1 ) d) ]

→5/2 ( 2a + 4d) = -((10/2)(2a + 9d))

put the value of a

→ 5/2(2(14)+ 4d) = -((5 (2(14)+ 9d)))

→ 5/2 ( 28 + 4d) = -((5 (28 + 9d)))

→[(5/2)x (28) + (5/2) x (4d) ]

= - (( 140 + 45d))

→ ( 5 x 14 ) + ( 5 x 2d)

= - 140 - 45d

→ 70 + 10d = -140 - 45d

→ 70 + 140 = -45d - 10d

→ 210 = -55d

→ 210 / 55 = d

→ d = - 3.81

so the common difference d is

$\boxed{\textbf{\large{ d = -3.81}}}$

▶Therefor, we have to find the 3rd term of an A. P

we know, ★ tn = a +(n - 1) d

t3 = a + ( 3 - 1) d

put the values of a and d

→ t3 = 14 + 2 x (- 3.81 )

→ t3 = 14 + ( -7.62)

→ t3 = 14 - 7.62

→ t3 = 6.38

$\boxed{\textbf{\large{3rd term = 6.38}}}$