Question

The first term of an AP 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

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Answer 2

Answer:

$6 \frac{4}{11}$

Step-by-step explanation:

a=14

S5= -S10

So,

5/2 [2(14)+4d] = -10/2 [2(14)+9d]

cancelling the 2 from denominator, we get

5(28+4d) = -10(28+9d)

28+4d = -2(28+9d)

28+4d = -56-18d

22d = -84

d= -84/2 = -42/11

now,

third term,

since

tn= a+(n-1)d

therefore,

t3= 14+(3-1) × (-42/11)

t3 = 14+ (2× -42/11)

t3 = 14-84/2

t3 = 154-84/11

t3 = 70/11 =

$6 \frac{4}{11}$

HOPE THIS HELPS