Question

What is 5th term of an A.P. whose first term and common difference are 7 and -2 respectively.​

Answer 1

Step-by-step explanation:

We have,

Let first term is an A.P.

a=−14

Fifth term is a

5

=a+4d=2

Put a=−14 and we get,

a+4d=2

−14+4d=2

4d=2+14

4d=16

d=4

Find the value of n

Whose sum is 40.

Then,

We know that,

S

n

=

2

n

(2a+(n−1)d)

⇒40=

2

n

(2×(−14)+(n−1)×4)

⇒80=n(−28+4n−4)

⇒80=n(−32+4n)

⇒4n(n−8)=80

⇒n(n−8)=20

⇒n

2

−8n−20=0

⇒n

2

−(10−2)n−20=0

⇒n

2

−10n+2n−20=0

⇒n(n−10)+2(n−10)=0

⇒(n−10)(n+2)=0

For

n+2=0

n=−2

It is not possible (negative)

For,

n−10=0

n=10

Hence, this is the answer.

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Answer 2

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If the common difference is taken as d.

Then, a5 = a + 4d

⟹ 2 = -14 + 4d

⟹ 2 + 14 = 4d

⟹ 4d = 16

⟹ d = 4

Next, we know that Sn = n/2[2a + (n − 1)d]

Where; a = first term for the given A.P.

d = common difference of the given A.P.

n = number of terms

Now, on substituting the values in Sn

⟹ 40 = n/2 n2[2(−14) + (n − 1)(4)]

⟹ 40 = n/2[−28 + (4n − 4)]

⟹ 40 = n/2[−32 + 4n]

⟹ 40(2) = – 32n + 4n2

So, we get the following quadratic equation,

4n2 – 32n – 80 = 0

⟹ n2 – 8n + 20 = 0

On solving by factorization method, we get

4n2 – 10n + 2n + 20 = 0

⟹ n(n – 10) + 2( n – 10 ) = 0

⟹ (n + 2)(n – 10) = 0

Either, n + 2 = 0

⟹ n = -2

Or, n – 10 = 0

⟹ n = 10

Since the number of terms cannot be negative.

Therefore, the number of terms (n) is 10.