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Question : 110 of 150
Marks:
+1,
-0
Solution:
Since,
32‌sin‌2‌α−1,14 and
34−2‌sin‌2‌α are in AP. Therefore,
2×14=32‌sin‌2‌α−1+34−2‌sin‌2‌α ⇒28=+, where
a=32‌sin‌2‌α ⇒a2−84a+243=0 ⇒(a−81)(a−3)=0 ⇒a=81,a=3 ⇒32‌sin‌2‌α=34 or
32‌sin‌2‌α=3 ⇒ 2‌sin‌2‌α=1‌‌‌‌[∵2‌sin‌2‌α≠4] ⇒
sin‌2‌α=⇒2α=30° [∵sin‌30°=1∕2] Thus, the first three terms of the
AP are
1,14,27.
Hence, its fifth term
a5=a1+(5−1)d =1+4×13=1+52=53
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