Question
The eighth term of an AP is half its second term and the eleventh term exceeds one third of its fourth term by 1. Find the $15^{th}$ term.
✓
Answer
Consider the first term and common difference as a and d respectively.
$\text{a}_{8}=\frac{1}{2}\text{a}_{2}[\text{Given}]$
[$\because$ $a_n = a + (n - 1)d]$
$\Rightarrow\text{a}+(8-1)\text{d}=\frac{1}{2}\big[\text{a}+\big(2-1)\text{d}\big]$
$2(a + 7d) = a + d 2a + 14d - a - d = 0 a + 13d = 0 ........ (i)$
$\text{Now},\text{ a}_{11}=\frac{1}{3}\text{a}_{4}+1\big[\text{Given}\big]$
$\Rightarrow\text{a}+\big(11-1\big)\text{d}=\frac{1}{3}\big[\text{a}+\big(4-1)\text{d}\big]+1$
$\Rightarrow\big(\text{a}+10\text{d}\big)=\frac{1}{3}\big(\text{a}+3\text{d}\big)+1$
$3(a + 10d) = a + 3d + 3 3a + 30d - a - 3d = 3 2a + 27d = 3 ...... (ii)$ Multiplaying (i) by 2,
we have 2a + 26d = 0 ........ (iii)
Now, subtraction (iii) from (ii), we get
Now, a + 13d = 0 [From (i)]
$\Rightarrow a + 13 \times 3 = 0$
$\Rightarrow a = -39$
Now, we know that an
$= a + (n - 1)d$
$\Rightarrow a_{15} = -39 + (15 - 1)3$
$\Rightarrow a_{15} = 3$
$\text{a}_{8}=\frac{1}{2}\text{a}_{2}[\text{Given}]$
[$\because$ $a_n = a + (n - 1)d]$
$\Rightarrow\text{a}+(8-1)\text{d}=\frac{1}{2}\big[\text{a}+\big(2-1)\text{d}\big]$
$2(a + 7d) = a + d 2a + 14d - a - d = 0 a + 13d = 0 ........ (i)$
$\text{Now},\text{ a}_{11}=\frac{1}{3}\text{a}_{4}+1\big[\text{Given}\big]$
$\Rightarrow\text{a}+\big(11-1\big)\text{d}=\frac{1}{3}\big[\text{a}+\big(4-1)\text{d}\big]+1$
$\Rightarrow\big(\text{a}+10\text{d}\big)=\frac{1}{3}\big(\text{a}+3\text{d}\big)+1$
$3(a + 10d) = a + 3d + 3 3a + 30d - a - 3d = 3 2a + 27d = 3 ...... (ii)$ Multiplaying (i) by 2,
we have 2a + 26d = 0 ........ (iii)
Now, subtraction (iii) from (ii), we get
Now, a + 13d = 0 [From (i)]
$\Rightarrow a + 13 \times 3 = 0$
$\Rightarrow a = -39$
Now, we know that an
$= a + (n - 1)d$
$\Rightarrow a_{15} = -39 + (15 - 1)3$
$\Rightarrow a_{15} = 3$
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