Question

in an A.P the third and the fifth terms are 6 and 10 respectively. find the 1.first term 2.common difference
3.sum of the first 12 terms
​

Answer 1

Answer:

1. a= 2

2. d= 2

3. Sum of first 12 terms = 156

Answer 2

To Find :-

• The First Term of the AP

• Common difference

• Sum of 12 terms

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

Given :-

• 3rd term of the AP = 6

• 5th term of the AP = 10

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

We know :-

⠀⠀⠀⠀⠀⠀Formula for nth term :-

$\boxed{\underline{\bf{t_{n} = a_{1} + (n -1)d}}}$

Where :-

• $t_{n}$ = nth term
• $a_{1}$ = First Term
• $d$ = Common Difference
• $n$ = no. of terms

⠀⠀⠀⠀⠀Formula for Sum of AP's :-

$\boxed{\underline{\bf{s_{n} = \dfrac{n}{2}\bigg[2a + (n - 1)d\bigg]}}}$

Where :-

• $s_{n}$ = Sum of AP's
• $a$ = First Term
• $n$ = no. of terms
• $d$ = Common Difference

Solution :-

Equation (i) :-

Given :-

• 3rd term = 6

Taken :-

• Let the first term be a.
• Let the Common Difference be d.

Using the formula and substituting the values in it, we get :-

$:\implies \bf{t_{n} = a_{1} + (n -1)d} \\ \\ \\ :\implies \bf{t_{3} = a_{1} + (n - 1)d} \\ \\ \\ :\implies \bf{6 = a_{1} + (3 -1)d} \\ \\ \\ :\implies \bf{6 = a_{1} + 2d} \\ \\ \\ \therefore \purple{\bf{6 = a_{1} + 2d}} [Equation.(i)]$

Hence, equation one is $\bf{6 = a_{1} + 2d}$.

Equation (ii) :-

Given :-

• 3rd term = 6

Taken :-

• Let the first term be a.
• Let the Common Difference be d.

Using the formula and substituting the values in it, we get :-

$:\implies \bf{t_{n} = a_{1} + (n -1)d} \\ \\ \\ :\implies \bf{t_{5} = a_{1} + (n - 1)d} \\ \\ \\ :\implies \bf{10 = a_{1} + (5 -1)d} \\ \\ \\ :\implies \bf{10 = a_{1} + 4d} \\ \\ \\ \therefore \purple{\bf{10 = a_{1} + 4d}} [Equation.(ii)]$

Hence, equation two is $\bf{10 = a_{1} + 4d}$

Putting the two Equations together , we get :-

⠀⠀⠀⠀⠀⠀⠀⠀⠀$\bf{6 = a_{1} + 2d}$

⠀⠀⠀⠀⠀⠀⠀⠀⠀$\bf{10 = a_{1} + 4d}$⠀⠀⠀⠀⠀⠀⠀⠀⠀_______________

⠀⠀⠀⠀⠀⠀⠀⠀⠀$\bf{-4 = -2d}$

⠀⠀⠀⠀⠀⠀⠀⠀⠀

⠀⠀⠀⠀⠀$:\implies \bf{\not{-}4 = \not{-}2d}$

⠀⠀⠀⠀⠀$:\implies \bf{4 = 2d}$⠀

⠀⠀⠀⠀⠀$:\implies \bf{\dfrac{4}{2} = d}$

⠀⠀⠀⠀⠀$:\implies \bf{2 = d}$

⠀⠀⠀⠀⠀$\therefore \purple{\bf{a = 2}}$

Hence, the common difference is 2.

Now , putting the value of common difference in the Equation (i) , we get :-

$:\implies \bf{6 = a_{1} + 2d} \\ \\ :\implies \bf{6 = a_{1} + 2 \times 2} \\ \\ :\implies \bf{6 = a_{1} + 4} \\ \\ :\implies \bf{6 - 4 = a_{1}} \\ \\ :\implies \bf{2 = a_{1}} \\ \\ \therefore \purple{\bf{a_{1} = 2}}$

Hence, the First Term of the AP is 2.

Sum of first 12 terms :-

Given :-

• First Term (a) = 2
• Common Difference (d) = 2
• Number of terms (n) = 12

Using the formula and substituting the values in it , we get :-

$:\implies \bf{s_{n} = \dfrac{n}{2}\bigg[2a + (n - 1)d\bigg]} \\ \\ \\ :\implies \bf{s_{12} = \dfrac{12}{2}\bigg[2 \times 2 + (12 - 1) \times 2\bigg]} \\ \\ \\ :\implies \bf{s_{12} = 6 \times (4 + 11 \times 2)} \\ \\ \\ :\implies \bf{s_{12} = 6 \times (4 + 22)} \\ \\ \\ :\implies \bf{s_{12} = 6 \times 26} \\ \\ \\ :\implies \bf{s_{12} = 156} \\ \\ \\ \therefore \purple{\bf{s_{12} = 156}}$

Hence, the sum of 12 terms is 156.

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