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Study Stuff for MCA Examinations - NIMCET
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Previous Year Question (PYQs)
Suppose $t_1, t_2, ...t_5$ are in AP such that $\sum ^{18}_{l=0}{{t}}_{3l+1}=1197$ and ${{t}}_7+{{3}}t_{22}=174$. If $\sum ^9_{l=1}{{{t}}_l}^2=947b$, then the value of $b$ is
Solution
Let first term $= a$, common difference $= d$, so $t_n = a + (n-1)d$Step 1: Expand $\displaystyle\sum_{l=0}^{18} t_{3l+1} = 1197$
Terms: $t_1, t_4, t_7, \ldots, t_{55}$ (19 terms, $l = 0$ to $18$)
$t_{3l+1} = a + 3ld$
$\displaystyle\sum_{l=0}^{18}(a + 3ld) $
$= 19a + 3d\cdot\dfrac{18 \times 19}{2} $
$= 19a + 513d = 1197$
$\Rightarrow a + 27d = 63 \quad \cdots(1)$
Step 2: Use $t_7 + 3t_{22} = 174$
$t_7 = a + 6d,\quad t_{22} = a + 21d$
$(a + 6d) + 3(a + 21d) = 174$
$4a + 69d = 174 \quad \cdots(2)$
Step 3: Solve (1) and (2)
From (1): $a = 63 - 27d$
Substitute in (2):
$4(63 - 27d) + 69d = 174$
$252 - 108d + 69d = 174$
$-39d = -78$
$d = 2$
$a = 63 - 54 = 9$
Step 4: Find $\displaystyle\sum_{l=1}^{9} t_l^2$
$t_l = 9 + (l-1) \times 2 = 7 + 2l$
$\displaystyle\sum_{l=1}^{9} t_l^2 = \sum_{l=1}^{9}(7+2l)^2 = \sum_{l=1}^{9}(49 + 28l + 4l^2)$
$= 49(9) + 28\cdot\dfrac{9 \times 10}{2} + 4\cdot\dfrac{9 \times 10 \times 19}{6}$
$= 441 + 1260 + 1140$
$= 2841$
Step 5: Find $b$
$947b = 2841$
$b = \dfrac{2841}{947} = 3$
Answer: $b = \boxed{3}$
$\Rightarrow a + 27d = 63 \quad \cdots(1)$
Step 2: Use $t_7 + 3t_{22} = 174$
$t_7 = a + 6d,\quad t_{22} = a + 21d$
$(a + 6d) + 3(a + 21d) = 174$
$4a + 69d = 174 \quad \cdots(2)$
Step 3: Solve (1) and (2)
From (1): $a = 63 - 27d$
Substitute in (2):
$4(63 - 27d) + 69d = 174$
$252 - 108d + 69d = 174$
$-39d = -78$
$d = 2$
$a = 63 - 54 = 9$
Step 4: Find $\displaystyle\sum_{l=1}^{9} t_l^2$
$t_l = 9 + (l-1) \times 2 = 7 + 2l$
$\displaystyle\sum_{l=1}^{9} t_l^2 = \sum_{l=1}^{9}(7+2l)^2 = \sum_{l=1}^{9}(49 + 28l + 4l^2)$
$= 49(9) + 28\cdot\dfrac{9 \times 10}{2} + 4\cdot\dfrac{9 \times 10 \times 19}{6}$
$= 441 + 1260 + 1140$
$= 2841$
Step 5: Find $b$
$947b = 2841$
$b = \dfrac{2841}{947} = 3$
Answer: $b = \boxed{3}$
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