Let the $n$-th term be the first negative term: $a_n < 0$
Formula: $a + (n - 1)d < 0$
$20 + (n - 1)\left(-\frac{3}{4}\right) < 0$
$20 < (n - 1)\left(\frac{3}{4}\right)$
$80 < 3(n - 1)$
$3n - 3 > 80 \implies 3n > 83$
$n > \frac{83}{3} \implies n > 27.67$
Since $n$ must be the next integer greater than $27.67$, $n = 28$.
Q20. Find the number of terms in the finite AP: $7, 13, 19, \dots, 205$.
Answer: The number of terms in the AP is $34$. Result:
First term ($a$) = $7$
Common difference ($d$) = $13 - 7 = 6$
Last term ($a_n$) = $205$
Formula: $a + (n - 1)d = a_n$
$7 + (n - 1)6 = 205$
$(n - 1)6 = 205 - 7 = 198$
$n - 1 = \frac{198}{6} = 33$
$n = 33 + 1 = 34$
Q21. If 4 times the 4th term of an AP is equal to 18 times its 18th term, find its 22nd term.
Answer: The 22nd term of the AP is $0$. Result:
General term formula: $a_n = a + (n - 1)d$
Given condition: $4 \times a_4 = 18 \times a_{18}$ (Note: Typically standard textbook problems state "18 times the 18th term" or similar, let's set up based on the literal text: $4a_4 = 18a_{18}$)
Substitute expressions: $4(a + 3d) = 18(a + 17d)$
$4a + 12d = 18a + 306d$
Bring all terms to one side: $18a - 4a + 306d - 12d = 0$
$14a + 294d = 0$
Divide the entire equation by $14$: $a + 21d = 0$
Since $a_{22} = a + (22 - 1)d = a + 21d$, it follows that $a_{22} = 0$.
Q22. Find the middle term of the AP:
$6, 13, 20, \dots, 216$.
Answer: The middle term of the AP is $111$. Result:
First term ($a$) = $6$
Common difference ($d$) = $13 - 6 = 7$
Last term ($a_n$) = $216$
Find the total number of terms ($n$): $a + (n - 1)d = a_n$
$6 + (n - 1)7 = 216$
$(n - 1)7 = 216 - 6 = 210$
$n - 1 = \frac{210}{7} = 30$
$n = 30 + 1 = 31$
Since $n = 31$ is odd, there is one middle term, which is the $\left(\frac{31 + 1}{2}\right)$-th term, i.e., the 16th term.
Find the 16th term ($a_{16}$): $a_{16} = a + 15d = 6 + 15(7)$
$a_{16} = 6 + 105 = 111$
Q23. The sum of the 4th and 8th terms of an AP is 24 and the sum of the 6th and 10th terms is 44. Find the first term.
Answer: The first term of the AP is $-13$. Result:
Subtract Equation 1 from Equation 2: $(a + 7d) - (a + 5d) = 22 - 12$
$2d = 10 \implies d = 5$
Substitute $d = 5$ into Equation 1: $a + 5(5) = 12 \implies a + 25 = 12$
$a = 12 - 25 = -13$
Q24. Determine the AP whose third term is 16 and the 7th term exceeds the 5th term by 12.
Answer: The AP is $4, 10, 16, 22, \dots$ Result:
General term formula: $a_n = a + (n - 1)d$
Given third term ($a_3 = 16$):
$a + 2d = 16$ --- (Equation 1)
Given that the 7th term exceeds the 5th term by 12 ($a_7 - a_5 = 12$):
$(a + 6d) - (a + 4d) = 12$
$2d = 12 \implies d = 6$
Substitute $d = 6$ into Equation 1: $a + 2(6) = 16 \implies a + 12 = 16 \implies a = 4$
Construct the AP using $a = 4$ and $d = 6$:
$a_1 = 4$
$a_2 = 4 + 6 = 10$
$a_3 = 10 + 6 = 16$
$a_4 = 16 + 6 = 22$
Thus, the AP is $4, 10, 16, 22, \dots$
Q25. For what value of $n$, are the $n^{\text{th}}$ terms of two APs: $63, 65, 67, \dots$ and $3, 10, 17, \dots$ equal?
Answer: The $n^{\text{th}}$ terms are equal when $n = 13$. Result:
General term formula: $a_n = a + (n - 1)d$
For the first AP ($63, 65, 67, \dots$):
First term ($a_1$) = $63$, Common difference ($d_1$) = $65 - 63 = 2$
$n^{\text{th}}$ term of the first AP = $63 + (n - 1)2$
For the second AP ($3, 10, 17, \dots$):
First term ($a_2$) = $3$, Common difference ($d_2$) = $10 - 3 = 7$
$n^{\text{th}}$ term of the second AP = $3 + (n - 1)7$
Equating both $n^{\text{th}}$ terms:
$63 + (n - 1)2 = 3 + (n - 1)7$
$63 - 3 = (n - 1)7 - (n - 1)2$
$60 = (n - 1)(7 - 2)$
$60 = (n - 1)5$
$n - 1 = \frac{60}{5} = 12$
$n = 12 + 1 = 13$
Academic Repository Disclaimer & எச்சரிக்கை அறிவிப்பு :
English: These model solutions are structured to global academic standards for self-assessment and practice purposes. Students are advised to verify with official textbooks for board examinations.
தமிழ்: இந்த மாதிரி கணிதத் தீர்வுகள் மாணவர்களின் சுய கற்றல் மற்றும் பயிற்சித் திறனுக்காக உலகளாவிய கல்வித் தரத்தில் உருவாக்கப்பட்டுள்ளன. தேர்வுகளுக்குத் தயாராகும் போது உங்கள் அதிகாரப்பூர்வ பாடப்புத்தகத்துடன் சரிபார்க்கவும்.
