Question

Two APs have the same common difference. The first term of one of these is -1 and that of the other is -8. Then the difference between their 4th terms is a) -1 b) -8 c) 7 d) 9

Answer 1

GIVEN :–

• Two A.P.'s have the same common difference.

• The first term of one of these is -1 and that of the other is -8.

TO FIND :–

• Difference between their 4th terms is = ?

SOLUTION :–

• We know that –

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$\bf \implies T_n =a+(n-1)d$

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• For first A.P. ( 4th term ) –

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$\bf \: \: \: {\huge{.}} \: \: \: a = - 1$

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$\bf \: \: \: {\huge{.}} \: \: \: common \: \: difference =d$

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$\bf \: \: \: {\huge{.}} \:\:\:n=4$

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$\bf \: \: \: {\huge{.}} \:\:\:T_n = T_n$

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• So that –

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$\bf \implies T_4 =( - 1)+(4 -1)d$

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$\bf \implies T_4 =( - 1)+(3)d$

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$\bf \implies T_4 =3d - 1$

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• For second A.P.( 4th term ) –

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$\bf \: \: \: {\huge{.}} \: \: \: a = - 8$

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$\bf \: \: \: {\huge{.}} \: \: \: common \: \: difference =d$

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$\bf \: \: \: {\huge{.}} \:\:\:n=4$

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$\bf \: \: \: {\huge{.}} \:\:\:T_n = T'_n$

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• So that –

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$\bf \implies T'_4 =( - 8)+(4 -1)d$

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$\bf \implies T'_4 =( - 8)+(3)d$

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$\bf \implies T'_4 =3d - 8$

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• Now difference –

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$\bf \implies T_4 - T'_4 =(3d - 1) - (3d - 8)$

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$\bf \implies T_4 - T'_4 =3d - 1- 3d + 8$

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$\bf \implies T_4 - T'_4 =8 - 1$

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$\bf \implies \large{ \boxed{ \bf T_4 - T'_4 =7}}$

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▪︎Hence, Option (c) is correct.

Answer 2

Answer:

c) 7

Step-by-step explanation:

a4 = a'4

an = a + (n - 1)d

a + (4 - 1)d = a' + (4 - 1)d

a + 3d = a' + 3d

Given that, the first term of one of these is -1 and that of the other is -8.

→ a4 - a'4

→ -1 + 3d - [(-8) + 3d]

→ -1 + 3d + 8 - 3d

→ 7

Hence, the difference between their 4th terms is 7.