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प्रश्न
Find the 22nd term of the progression `7 3/4 + 9 1/2 + 11 1/4 + ...`
योग
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उत्तर
a1 = `7 3/4`
= `31/4`
d = `9 1/2 − 7 3/4`
d = `9 (1 xx 2)/(2 xx 2) - 7 3/4`
d = `9 2/4 - 7 3/4`
d = `38/4 - 31/4`
d = `7/4`
a22 = a1 + 21 × d
= `31/4 + 21 xx (7/4)`
= `(31 + 147)/4`
= `178/4 = 89/2`
= `44 1/2`
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