Question

10. Find the first term and the common difference of the A.P.
when
T4 = 7 and T16 = 31​

Answer 1

Answer:-

$\sf T_4 = 7$

$\sf a+(n-1) d= 7$

$\sf a +(4-1) d = 7$

$\sf a + 3d =7 \: \: \: \: \: - - - - - (i)$

$\sf T_{16} = 31$

$\sf a + (16-1)=31$

$\sf a +15 d = 31 \: \: \: \: \: - - - - (ii)$

• Subtracting equation (I) to (ii)

we get,

$\sf 12d=4$

$\sf \implies \: d = \cancel\frac{4}{12}$

$\sf \implies \: d = \frac{1}{3}$

• Putting the value of d = 1/3 in (I)

$\sf a+3d=7$

$\sf \implies \: a+ 3× \frac{1}{3} = 7$

$\sf \implies \: a = 7$

• a = 7
• a = 7d = 1/3

The first term is 7 and the common difference is 1/3 of the A.P.

Answer 2

Answer:

The first term and the common difference of the AP are 1 & 2 respectively.

Step-by-step-explanation:

We have given that, for an AP,

t₄ = 7

t₁₆ = 31

We have to find the first term and common difference of AP.

Let "a" be the first term and "d" be the common difference of the AP.

We know that,

tₙ = a + ( n - 1 ) * d - - - [ Formula ]

∴ t₄ = a + ( 4 - 1 ) * d

⇒ 7 = a + 3 * d

⇒ 7 = a + 3d

⇒ a = 7 - 3d - - - ( 1 )

Now,

t₁₆ = a + ( 16 - 1 ) * d

⇒ 31 = a + 15 * d

⇒ 31 = a + 15d

⇒ a + 15d = 31

⇒ 7 - 3d + 15d = 31 - - - [ From ( 1 ) ]

⇒ 12d = 31 - 7

⇒ 12d = 24

⇒ d = 24 ÷ 12

⇒ d = 2

By substituting d = 2 in equation ( 1 ), we get,

a = 7 - 3d

⇒ a = 7 - 3 * 2

⇒ a = 7 - 6

⇒ a = 1

∴ The first term and the common difference of the AP are 1 & 2 respectively.