Question

1. If 7 times the seventh term of an AP is 11 times the 11th term, then its 18th

term will be

a. 7 b. 11 c. 18 d.0 ​

Answer 1

$\large\bf\underline {To \: find:-}$

• we need to find the 18th term of AP

$\huge\bf\underline{Solution:-}$

$\bf\underline{\red{Given:-}}$

• 7 times the the seventh term of an AP is 11 times the 11th term.

$\sf \: 7a_7 = 11a_{11}$

we know that,

$\star \bf \large \: a_n = a + (n-1)d$

• a7 = a + 6d
• a11 = a + 10d

→ 7(a + 6d )= 11(a + 10d)

→ 7a + 42d = 11a + 110d

→ 7a - 11a = 110d - 42d

→ - 4a = 68d

→ 4a = - 68d

→ a = -68/4

→ a = - 17d ......1)

• Now, finding 18th term of AP:-

→ a18 = a + 17d

• From equation (1)

→ a18 = - 17d + 17d

→ a18 = 0

Hence,

• 18th term of AP is 0

So,

• Option d) is correct.✓

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Answer 2

$\sf\huge{\green{\underbrace{\orange{To\:Find:-}}}}$

• 18th term .

$\sf\huge{\underbrace{\red{SOLUTION:-}}}$

FORMULA :-

☞ For ‘n’th terms,

$\bigstar\:\tt{\underline{\purple{\boxed{a_{n^{th}}\:=\:a\:+\:(n-1)\:d\:}}}}$

• a = First term

• d = common difference

• n = term number

GIVEN :-

• 7 times the 7th term of an A.P is equal to 11 times the 11th term .

CALCULATION :-

1) First, calculate the 7th term of A.P .

$\tt{a_{7th}\:=\:a\:+\:(7-1)\:d\:}$

$\tt{\implies\:a_{7th}\:=\:a\:+\:6d\:}$ -----(1)

2) Secondly, calculate the 11th term of A.P .

$\tt{a_{11th}\:=\:a\:+\:(11-1)\:d\:}$

$\tt{\implies\:a_{11th}\:=\:a\:+\:10d\:}$ -----(2)

☞ Now, According to question

$\tt{\blue{\boxed{7(a_{7th})\:=\:11(a_{11th})\:}}}$

$\tt{\implies\:7(a\:+\:6d)\:=\:11(a\:+\:10d)\:}$

$\tt{\implies\:7a\:+\:42d\:=\:11a\:+\:110d\:}$

$\tt{\red{\boxed{\implies\:a\:=\:(-17d)\:}}}$

☞ Now, 18th term of A.P is,

$\tt{a_{18th}\:=\:a\:+\:(18-1)\:d\:}$

$\tt{\implies\:a_{18th}\:=\:a\:+\:17d\:}$

➜ Putting the value of “a = (-17d)” in the above equation, we get

$\tt{\implies\:a_{18th}\:=\:(-17d)\:+\:17d\:}$

$\tt{\green{\boxed{\implies\:a_{18th}\:=\:0\:}}}$ -----(d) .

__________________________

$\bigstar\:\underline{\boxed{\bf{\purple{Required\:Answer\::\:[a_{18th}\:=\:0]\:---(d)\:}}}}$