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महाराष्ट्र राज्य शिक्षण मंडळएस.एस.सी (इंग्रजी माध्यम) इयत्ता १० वी

T_19 = ? for the given A.P., 9, 4, –1, –6...... Activity :- Here a = 9, d = □ t_n = a + (n – 1)d t_19 = 9 + (19 – 1) □ = 9 + □ = □

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प्रश्न

t19 = ? for the given A.P., 9, 4, –1, –6......

Activity :- Here a = 9, d = `square`

tn = a + (n – 1)d

t19 = 9 + (19 – 1) `square`

= 9 + `square`

= `square`

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उत्तर

Here a = 9, d = 4 – 9 = \[\boxed{-5}\]

tn = a + (n – 1)d

t19 = 9 + (19 – 1) \[\boxed{-5}\]

= 9 + 18(– 5)

= 9 + \[\boxed{-90}\]

= \[\boxed{-81}\] 

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पाठ 3: Arithmetic Progression - Q.2 (A)
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