Question

How many terms are there in AP whose first terms and 6 terms are _12 and 8 respectively and sum of all its terms 120

Answer 1

Answer:

-4272

Step-by-step explanation:

t1 =a =12

t6= 8

t6 =a + ( n-1) d

8= 12 + (6-1)d

d= -4/5

S120 = N/2 [2a +(n-1) d]

= 120/2 [ 2× 12 + ( 120-1) -4/5 ]

= - 4272

Answer 2

The sum will be "S₁₂₀ = 4272". Further explanation is given below.

Step-by-step explanation:

Given:

The first term, T₁ , a = 12

Sixth term, T₆ = 8

As we know,

Tn = a+(n-1)d

On putting the value of T₆, a, n and d, we get

⇒  8 = 12+(6-1)d

⇒  8 = 12+5d

⇒  8-12 = 5d

⇒  -4 = 5d

⇒   d = $\frac{-4}{5}$

Also, S₁₂₀ = [2a+(n-1)d]

On putting the values in the above formula, we get

⇒  S₁₂₀ = [2×12+(120-1)$\frac{-4}{5}$]

⇒         = 60[24+(119×($\frac{-4}{5}$))]

⇒         = 60[24-95.2)]

⇒         = 60(-71.2)

⇒         = - 4272

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sum of all 120?...

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