Question

The 7th term of an arithmetic sequence is 20 and the 15th te is -4 .what is the common difference ​

Answer 1

EXPLANATION.

7th term of an A.P. = 20.

15th term of an A.P. = - 4.

As we know that,

Formula of :

General term of an A.P.

⇒ Tₙ = a + (n - 1)d.

Using this formula in the equation, we get.

⇒ T₇ = 20.

⇒ T₇ = a + (7 - 1)d.

⇒ T₇ = a + 6d.

⇒ a + 6d = 20. - - - - - (1).

⇒ T₁₅ = - 4.

⇒ T₁₅ = a + (15 - 1)d.

⇒ T₁₅ = a + 14d.

⇒ a + 14d = - 4. - - - - - (2).

From equation (1) and (2), we get.

⇒ a + 6d = 20. - - - - - (1).

⇒ a + 14d = - 4. - - - - - (2).

Subtract equation (1) and (2), we get.

⇒ a + 6d = 20. - - - - - (1).

⇒ a + 14d = - 4. - - - - - (2).

⇒ -  -           +

We get,

⇒ - 8d = 24.

⇒ d = - 3.

Put the value of d = - 3 in the equation (1), we get.

⇒ a + 6d = 20.

⇒ a + 6(-3) = 20.

⇒ a - 18 = 20.

⇒ a = 20 + 18.

⇒ a = 38.

Sequences will be written as,

⇒ (a), (a + d), (a + 2d), (a + 3d), . . . . .

⇒ (38), (38 + (-3)), (38 + 2(-3)), (38 + 3(-3)), . . . . .

⇒ (38), (38 - 3), (38 - 6), (38 - 9), . . . . .

⇒ (38), (35), (32), (29), . . . . .

First term = a = 38.

Common difference = d = - 3.

                                                                                                                     

MORE INFORMATION.

Supposition of terms in an A.P.

(1) = Three terms as : a - d, a, a + d.

(2) = Four terms as : a - 3d, a - d, a + d, a + 3d.

(3) = Five terms as : a - 2d, a - d, a, a + d, a + 2d.

Answer 2

Answer:

Appropriate Question :-

• The 7th term of an arithmetic sequence is 20 and the 15th term is - 4. What is the common difference (d) and the first term (a).

Given :-

• The 7th term of an arithmetic sequence is 20 and the 15th term is - 4.

To Find :-

• What is the common difference and the first term.

Formula Used :-

$\clubsuit$ General Term (nth term) of an AP Formula :

$\mapsto \sf\boxed{\bold{\pink{a_n =\: a + (n - 1)d}}}$

where,

• $\sf a_n$ = nth term of an AP
• a = First term of an AP
• n = Number of terms of an AP
• d = Common difference of an AP

Solution :-

❒ In case of 7th term of an AP :

$\implies \sf a_7 =\: a + (7 - 1)d$

$\implies \sf a_7 =\: a + (6)d$

$\implies \sf a_7 =\: a + 6d$

$\implies \sf 20 =\: a + 6d$

$\implies \sf\bold{\purple{a + 6d =\: 20\: ------\: (Equation\: No\: 1)}}$

❒ In case of 15th term of an AP :

$\implies \sf a_{15} =\: a + (15 - 1)d$

$\implies \sf a_{15} =\: a + (14)d$

$\implies \sf a_{15} =\: a + 14d$

$\implies \sf - 4 =\: a + 14d$

$\implies \sf\bold{\purple{a + 14d =\: - 4\: ------\: (Equation\: No\: 2)}}$

From the equation no 1 and 2 we get,

$\implies \sf a + 6d - (a + 14d) =\: 20 - (- 4)$

$\implies \sf a + 6d - a - 14d =\: 20 + 4$

$\implies \sf {\cancel{a}} {\cancel{- a}} + 6d - 14d =\: 24$

$\implies \sf 6d - 14d =\: 24$

$\implies \sf - 8d =\: 24$

$\implies \sf d =\: \dfrac{\cancel{24}}{- \cancel{8}}$

$\implies \sf d =\: \dfrac{3}{- 1}$

$\implies \sf\bold{\red{d =\: - 3}}$

Now, again by putting the value of d = - 3 in the equation no 2 we get,

$\implies \sf a + 14d =\: - 4$

$\implies \sf a + 14(- 3) =\: - 4$

$\implies \sf a + 14 \times (- 3) =\: - 4$

$\implies \sf a + (- 42) =\: - 4$

$\implies \sf a - 42 =\: - 4$

$\implies \sf a =\: - 4 + 42$

$\implies \sf\bold{\red{a =\: 38}}$

${\footnotesize{\bold{\underline{\therefore\: The\: common\: difference\: (d)\: and\: the\: first\: term\: (a)\: of\: an\: AP\: is\: - 3\: and\: 38\: respectively\: .}}}}$

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★ EXTRA INFORMATION ★

$\clubsuit$ Sum Of First n Term of an AP Formula :

$\mapsto \sf\boxed{\bold{\pink{S_n =\: \dfrac{n}{2}\bigg\lgroup 2a + (n - 1)d\bigg\rgroup}}}$

where,

• $\sf S_n$ = Sum of first n term of an AP
• n = Number of terms of an AP
• a = First term of an AP
• d = Common difference of an AP